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For this Question: A Tribonacci number T(n) is the sum of the preceding three elements in a series.
Calculate the value of the nth Tribonacci number modulo 1000000007. The initial three numbers of this series will be a, b and c i.e., T(1)=a, T(2)=b, T(3)=c.The only line of input contains 4 integers n, a, b, c.
Constraints
4 ≤ n ≤ 10^5
1 ≤ a, b, c ≤ 1000000000Print the result containing one integer i.e., T(n) modulo 1000000007.Input1:
5 1 2 3
Output1:
11
Input2:
4 1 1 1
Output2:
3
Explanation(might now be the optimal solution):
Input 1:
Follow the below steps:-
T(1) = 1
T(2) = 2
T(3) = 3
T(4) = 6
T(5) = 11, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main
{
static int mod = 1000000007;
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
//int t = Integer.parseInt(read.readLine());
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
long a =Long.parseLong(str[1]);
long b = Long.parseLong(str[2]);
long c = Long.parseLong(str[3]);
long dp[] = new long[n];
dp[0] = a;
dp[1] = b;
dp[2] = c;
for(int i = 3; i < n; i++)
{
dp[i] = (dp[i-1]%mod + dp[i-2]%mod + dp[i-3]%mod)%mod;
}
System.out.println(dp[n-1]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A Tribonacci number T(n) is the sum of the preceding three elements in a series.
Calculate the value of the nth Tribonacci number modulo 1000000007. The initial three numbers of this series will be a, b and c i.e., T(1)=a, T(2)=b, T(3)=c.The only line of input contains 4 integers n, a, b, c.
Constraints
4 ≤ n ≤ 10^5
1 ≤ a, b, c ≤ 1000000000Print the result containing one integer i.e., T(n) modulo 1000000007.Input1:
5 1 2 3
Output1:
11
Input2:
4 1 1 1
Output2:
3
Explanation(might now be the optimal solution):
Input 1:
Follow the below steps:-
T(1) = 1
T(2) = 2
T(3) = 3
T(4) = 6
T(5) = 11, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e3+ 5;
const int mod = 1e9 + 7;
const int inf = 1e18 + 9;
void solve(){
int n, a, b, c;
cin >> n >> a >> b >> c;
for(int i = 4; i <= n; i++){
int x = (a + b + c) % mod;
a = b;
b = c;
c = x;
}
cout << c << endl;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: def partition(array, low, high):
pivot = array[high]
i = low - 1
for j in range(low, high):
if array[j] <= pivot:
i = i + 1
(array[i], array[j]) = (array[j], array[i])
(array[i + 1], array[high]) = (array[high], array[i + 1])
return i + 1
def quick_sort(array, low, high):
if low < high:
pi = partition(array, low, high)
quick_sort(array, low, pi - 1)
quick_sort(array, pi + 1, high)
t=int(input())
for i in range(t):
n=int(input())
a=input().strip().split()
a=[int(i) for i in a]
quick_sort(a, 0, n - 1)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] quickSort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
public static int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: function quickSort(arr, low, high)
{
if(low < high)
{
let pi = partition(arr, low, high);
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
function partition(arr, low, high)
{
let pivot = arr[high];
let i = (low-1); // index of smaller element
for (let j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
let temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
int n;
while(t>0) {
n = Integer.parseInt(br.readLine());
int a[] = new int[n];
String s = br.readLine();
String arr[] = s.split(" ");
for(int i=0;i<n;i++) {
a[i] = Integer.parseInt(arr[i]);
}
int b[] = sort(a);
for(int i=0;i<n;i++) {
System.out.print(b[i] + " ");
}
System.out.println();
t--;
}
}
static int[] sort(int[] a) {
int temp, index;
for(int i=0;i<a.length-1;i++) {
index = i;
for(int j=i+1;j<a.length;j++) {
if(a[j]<a[index]) {
index = j;
}
}
temp = a[i];
a[i] = a[index];
a[index] = temp;
}
return a;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, I have written this Solution Code: for _ in range(int(input())):
n = input()
res = map(str, sorted(list( map(int,input().split()))))
print(' '.join(res)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: a, b, v = map(int, input().strip().split(" "))
c = abs(a-b)
t = c//v
print(t), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
System.out.print(Time(n,m,k));
}
static int Time(int A, int B, int S){
if(B>A){
return (B-A)/S;
}
return (A-B)/S;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2D grid. You are given two integers X and Y.
Consider a set "S" of points (x, y) on the cartesian plane such that 1 <= x <= N and 1 <= y <=M where x, y are positive integers.
You need to find number of line segments whose end points lies in set S such that the coordinates of the mid point also lies in the set S.The first line contains one integers – T (number of test cases).
The next T lines contains two integers N, M.
<b> Constraints: </b>
1 ≤ T ≤ 1000
1 ≤ N, M ≤ 1000Output T lines each containg a single integer denoting the number of such line segments.INPUT
2
3 3
6 7
OUTPUT
8
204, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
using cd = complex<double>;
const double PI = acos(-1);
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifndef ONLINE_JUDGE
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
ll nc2(ll n){
return (n*(n-1))/2;
}
void solve(){
ll x,y; cin >> x >> y;
ll a = x/2, b = (x+1)/2;
ll c = y/2, d = (y+1)/2;
// trace(a,b,c,d);
ll ans = 2* (nc2(a) + nc2(b)) * (nc2(c) + nc2(d));
// trace(ans);
ans += (x)*(nc2(c) + nc2(d));
// trace(ans);
ans += (nc2(a) + nc2(b))*y;
// trace(ans);
cout << ans << "\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ll t = 1;
cin >> t;
while(t--){
solve();
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The call stack of javascript based on the LIFO format i. e. last in first out.
Complete the functions <b>greetEnglish</b> and <b>greetFrench</b> so as the greetings are printed in the order : Hindi, English and French.Name of the person as stringGreetings are printed in the order : Hindi, English, French<b>Sample Input :</b>
John
<b>Sample Output :</b>
Namaste John!
Hello John!
Bonjour John!, I have written this Solution Code: function greetHindi(person) {
console.log(`Namaste ${person}!`)
}
function greetEnglish(person) {
greetHindi(person)
console.log(`Hello ${person}!`)
}
function greetFrench(person) {
greetEnglish(person)
console.log(`Bonjour ${person}!`)
}
function Greetings(person) {
greetFrench(person)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: def MaximumProduct(N):
ans=N//4
if N %4 !=0:
ans=ans+1
return ans, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given integer N, find the number of minimum elements the N has to be broken into such that their product is maximum.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MaximumProduct()</b> that takes integer N as parameter.
Constraints:-
1 <= N <= 10000Return the minimum number of elements the N has to be broken intoSample Input:-
5
Sample output
2
Explanation:-
N has to be broken into 2 and 3 to get the maximum product 6.
Sample Input:-
7
Sample Output:-
2
Explanation:- 4 + 3, I have written this Solution Code: static int MaximumProduct(int N){
int ans=N/4;
if(N%4!=0){ans++;}
return ans;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N boxes that are kept in a straight line. You are also given M colors such that ( M ≤ N
). You cannot change the position of boxes. Determine the number of ways to color the boxes such that if you select any M consecutive boxes then the color of each box is unique.
Since the number could be large, print the answer modulo 10<sup>9</sup> + 7
.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>CountBoxes()</b> that takes the integer N and M as parameter denoting the number of boxes and number of colours respectively.
<b>Constraints:</b>
1 ≤ M, N ≤ 10<sup>6</sup>Print the required answer in a new lineSample Input:-
2 2
Sample Output:-
2
Sample Input:-
4 3
Sample Output:-
6, I have written this Solution Code: class Solution {
public static void CountBoxes(int n,int m) {
long p = (long) (1e9 + 7);
long a = 1;
while (m > 0)
{
a = (a * m) % p;
m--;
}
System.out.print(a);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the minimum number of coins/ notes required to make the change of A amount of money.
For this problem, you can assume that there is an unlimited supply of the various notes/coins available in the Indian currency denominations. The various denominations available are 1, 2, 5, 10, 20, 50, 100, 200, 500 and 2000.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>minimumCoins()</b> which takes the target amount as a parameter.
Constraints:
1 <= Target < = 100000Return the minimum number of coins required.Sample Input:-
90
Sample Output:-
3
Explanation:-
50 + 20 + 20 = 90
Sample Input:-
2058
Sample Output:-
5
Explanation:-
2000 + 50 + 5 + 2 + 1, I have written this Solution Code: static int minimumCoins(int Target){
int ans=0;
int a[] = {1, 2, 5, 10, 20, 50, 100, 200, 500 , 2000};
for(int i=9;i>=0;i--){
ans+=Target/a[i];
Target=Target%a[i];
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. A string is called good if it is a prefix of string S. A string is called well if it is of even length and can be divided into two equal sub-strings of equal length. Find number of sub-strings which are both good and well.First line contains a string S.
Constraints
1<=|S|<=100000
String S can contain char (a-z, A-Z), or one digit integers (0-9).Print the answer.Sample Input
aacaacc
Sample Output
2
Explanation
aa, aacaac are two substrings
Sample Input
1111
Sample Output
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
int count=0;
int mid;
boolean flag=true;
if(str.length() >=2){
for(int i=0,j=1;j<str.length();j+=2){
mid=(j+1)/2;
flag=true;
for(int l=0,m=mid;i<mid && m<j+1;l++,m++){
if(str.charAt(l)!=str.charAt(m)){
flag=false;
break;
}
}
if(flag){
++count;
}
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. A string is called good if it is a prefix of string S. A string is called well if it is of even length and can be divided into two equal sub-strings of equal length. Find number of sub-strings which are both good and well.First line contains a string S.
Constraints
1<=|S|<=100000
String S can contain char (a-z, A-Z), or one digit integers (0-9).Print the answer.Sample Input
aacaacc
Sample Output
2
Explanation
aa, aacaac are two substrings
Sample Input
1111
Sample Output
2, I have written this Solution Code: s=input()
a=''
count=0
for i in range(len(s)//2+1):
a+=s[i]
if a==s[i+1:i+len(a)+1]:
count+=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. A string is called good if it is a prefix of string S. A string is called well if it is of even length and can be divided into two equal sub-strings of equal length. Find number of sub-strings which are both good and well.First line contains a string S.
Constraints
1<=|S|<=100000
String S can contain char (a-z, A-Z), or one digit integers (0-9).Print the answer.Sample Input
aacaacc
Sample Output
2
Explanation
aa, aacaac are two substrings
Sample Input
1111
Sample Output
2, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
//#define int long long
#define sz 2000005
int z[sz];
signed main()
{
string s;
cin>>s;
int n=s.size();
int l=0,r=0;
int no=0;
for(int i=1;i<n;i++)
{
if(i<=r) z[i]=min(r-l+1,z[i-l]);
while(i+z[i]<n && s[z[i]]==s[i+z[i]]) z[i]++;
if(i+z[i]-1>r) r=i+z[i]-1,l=i;
if(s[i]!='0' && z[i]>=i && i<=n/2) no++;
}
cout<<no<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2-D array of binary integers of size NXM. The cell at i<sup>th</sup> row and j<sup>th</sup> column is denoted by (i, j).
The array is called Balanced if every cell of the array having 4 elements has a balanced neighborhood. A neighborhood is balanced if 4 neighbors can be divided into 2 groups with equal size and equal sum.
A cell at (i, j) has 4 neighbors (i-1, j), (i+1, j), (i, j-1), (i, j+1).The first line contains N and M.
Next N lines contain M integers each.
<b>Constraints</b>
2 ≤ N, M ≤ 1000
0 ≤ arr[i][j] ≤ 1Print "YES" if the given array is balanced, otherwise print "NO".Input:
3 4
0 1 0 0
1 1 0 1
0 0 1 1
Output:
NO
Explanation:
neighbors of (1, 1) => {1, 1, 0, 0} => 1+0 = 0+1 => balanced neighborhood
neighbors of (1, 2) => {0, 1, 1, 1} => no way to divide into two groups with equal sum and equal size => unbalanced
No other cell has 4 neighbors., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
int m=Integer.parseInt(in.next());
int a[][] = new int[n][m];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++){
a[i][j] = Integer.parseInt(in.next());
}
}
int f=1;
for(int i=1;i+1<n;i++){
for(int j=1;j+1<m;j++){
int sum=a[i+1][j] + a[i-1][j] + a[i][j+1] + a[i][j-1];
if(sum%2 != 0){
f=-1;
break;
}
}
if(f==-1)break;
}
if(f==1)out.print("YES");
else out.print("NO");
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: static boolean isArmstrong(int N)
{
int num = N;
int sum = 0;
while(N > 0)
{
int digit = N%10;
sum += digit*digit*digit;
N = N/10;
}
if(num == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: function isArmstrong(n) {
// write code here
// do no console.log the answer
// return the output using return keyword
let sum = 0
let k = n;
while(k !== 0){
sum += Math.pow( k%10,3)
k = Math.floor(k/10)
}
return sum === n
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine().toString();
int n=Integer.parseInt(s);
int ch=0;
if(n>0){
ch=1;
}
else if(n<0) ch=-1;
switch(ch){
case 1: System.out.println("Positive");break;
case 0: System.out.println("Zero");break;
case -1: System.out.println("Negative");break;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: n = input()
if '-' in list(n):
print('Negative')
elif int(n) == 0 :
print('Zero')
else:
print('Positive'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch (num > 0)
{
// Num is positive
case 1:
printf("Positive");
break;
// Num is either negative or zero
case 0:
switch (num < 0)
{
case 1:
printf("Negative");
break;
case 0:
printf("Zero");
break;
}
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testcase = Integer.parseInt(br.readLine());
for(int t=0;t<testcase;t++){
int num = Integer.parseInt(br.readLine().trim());
if(num==1)
System.out.println("No");
else if(num<=3)
System.out.println("Yes");
else{
if((num%2==0)||(num%3==0))
System.out.println("No");
else{
int flag=0;
for(int i=5;i*i<=num;i+=6){
if(((num%i)==0)||(num%(i+2)==0)){
System.out.println("No");
flag=1;
break;
}
}
if(flag==0)
System.out.println("Yes");
}
}
}
}catch (Exception e) {
System.out.println("I caught: " + e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
if number > 1:
i=2
while i*i<=number:
if (number % i) == 0:
print("No")
break
i+=1
else:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n,k;
cin>>n;
long x=sqrt(n);
int cnt=0;
vector<int> v;
for(long long i=2;i<=x;i++){
if(n%i==0){
cout<<"No"<<endl;
goto f;
}}
cout<<"Yes"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You take part in <b>N</b> quizzes and in the i<sup>th</sup> quiz you get a score of Q<sub>i</sub>. Your friends are very competitive with you and they ask you the strength of your quiz scores. Strength of an array is defined as the following:
The maximum growth Q<sub>j</sub> - Q<sub>i</sub> (Q<sub>j</sub> > Q<sub>i</sub>) between two quizzes i and j such that i < j and there is no quiz <b>k</b> such that <b>i < k < j</b> and <b>Q<sub>k</sub> > Q<sub>i</sub></b>.
<b>If there is no such pair of indexes, print -1. </b>
Print the strength of your quiz marks in order to impress your friends.First line contains a single integer N, the number of quizzes.
The second line contains N space seperated integers Q<sub>1</sub>, Q<sub>2</sub>,. , Q<sub>N</sub> the score in each quiz.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= Q<sub>i</sub> <= 10<sup>9</sup>Print the strength of your quiz marks.Sample Input:
6
7 10 7 2 1 8
Sample Output:
7
Explaination:
There is a growth of 7 from Q<sub>5</sub> (= 1) to Q<sub>6</sub>(= 8). There is no growth in the array greater than this., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int N=Integer.parseInt(br.readLine());
String[] str=br.readLine().split(" ");
int[] arr=new int[N];
Deque<Integer> dq=new LinkedList<Integer>();
for(int i=0;i<N;i++)
{
arr[i]=Integer.parseInt(str[i]);
}
int res;
for(int i=1;i<N;i++)
{
res=arr[i]-arr[i-1];
while(!dq.isEmpty() && dq.getLast()<res)
{
dq.removeLast();
}
dq.addLast(res);
}
System.out.print(dq.getFirst());
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You take part in <b>N</b> quizzes and in the i<sup>th</sup> quiz you get a score of Q<sub>i</sub>. Your friends are very competitive with you and they ask you the strength of your quiz scores. Strength of an array is defined as the following:
The maximum growth Q<sub>j</sub> - Q<sub>i</sub> (Q<sub>j</sub> > Q<sub>i</sub>) between two quizzes i and j such that i < j and there is no quiz <b>k</b> such that <b>i < k < j</b> and <b>Q<sub>k</sub> > Q<sub>i</sub></b>.
<b>If there is no such pair of indexes, print -1. </b>
Print the strength of your quiz marks in order to impress your friends.First line contains a single integer N, the number of quizzes.
The second line contains N space seperated integers Q<sub>1</sub>, Q<sub>2</sub>,. , Q<sub>N</sub> the score in each quiz.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= Q<sub>i</sub> <= 10<sup>9</sup>Print the strength of your quiz marks.Sample Input:
6
7 10 7 2 1 8
Sample Output:
7
Explaination:
There is a growth of 7 from Q<sub>5</sub> (= 1) to Q<sub>6</sub>(= 8). There is no growth in the array greater than this., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve(){
int n;
cin >> n;
vector<int> a(n);
for(auto &i : a) cin >> i;
int ans = -1;
stack<int> s;
for(int i = n - 1; i >= 0; i--){
while(s.size() > 0 && s.top() <= a[i]){
s.pop();
}
if(s.size()) ans = max(ans, s.top() - a[i]);
s.push(a[i]);
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have two integers, X and Y. Initially, they are equal to A and B, respectively. In one move, you can do exactly one of the following operations:
1. Set X = X+1
2. Set Y = Y+1.
3. Let d be any divisor of X, then set X = d
4. Let d be any divisor of Y, then set Y = d.
Find the minimum number of moves so that X = Y.The input consists of two space separated integers A and B.
Constraints:
1 ≤ A, B ≤ 10<sup>9</sup>Print a single integer, the minimum number of moves so that X = Y.Sample 1:
1 8
Output 1:
1
Explanation 1:
We do only one operation, we let Y = 1 by the fourth operation. Then X = Y = 1.
Sample 2:
5 5
Output 2:
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
String s[]=bu.readLine().split(" ");
int a=Integer.parseInt(s[0]),b=Integer.parseInt(s[1]);
int ans=2;
if(a==b) ans=0;
else if(a%b==0 || b%a==0 || a+1==b || b+1==a) ans=1;
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have two integers, X and Y. Initially, they are equal to A and B, respectively. In one move, you can do exactly one of the following operations:
1. Set X = X+1
2. Set Y = Y+1.
3. Let d be any divisor of X, then set X = d
4. Let d be any divisor of Y, then set Y = d.
Find the minimum number of moves so that X = Y.The input consists of two space separated integers A and B.
Constraints:
1 ≤ A, B ≤ 10<sup>9</sup>Print a single integer, the minimum number of moves so that X = Y.Sample 1:
1 8
Output 1:
1
Explanation 1:
We do only one operation, we let Y = 1 by the fourth operation. Then X = Y = 1.
Sample 2:
5 5
Output 2:
0, I have written this Solution Code: l=list(map(int, input().split()))
A=l[0]
B=l[1]
if A==B:
print(0)
elif A%B==0 or B%A==0 or abs(A-B)==1:
print(1)
else:
print(2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have two integers, X and Y. Initially, they are equal to A and B, respectively. In one move, you can do exactly one of the following operations:
1. Set X = X+1
2. Set Y = Y+1.
3. Let d be any divisor of X, then set X = d
4. Let d be any divisor of Y, then set Y = d.
Find the minimum number of moves so that X = Y.The input consists of two space separated integers A and B.
Constraints:
1 ≤ A, B ≤ 10<sup>9</sup>Print a single integer, the minimum number of moves so that X = Y.Sample 1:
1 8
Output 1:
1
Explanation 1:
We do only one operation, we let Y = 1 by the fourth operation. Then X = Y = 1.
Sample 2:
5 5
Output 2:
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
if (a == b) cout << 0;
else if (abs(a - b) == 1) cout << 1;
else if (a % b == 0 || b % a == 0) cout << 1;
else cout << 2;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: int LastDigit(int N){
return N%10;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: static int LastDigit(int N){
return N%10;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: int LastDigit(int N){
return N%10;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: def LastDigit(N):
return N%10, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given is a positive integer N, where none of the digits is 0. Let k be the number of digits in N. We want to make a multiple of 3 by erasing at least 0 and at most k−1 digits from N and concatenating the remaining digits without changing the order. Determine whether it is possible to make a multiple of 3 in this way. If it is possible, find the minimum number of digits that must be erased to make such a number.The input consists of a single integer N.
<b>Constraints</b>
1≤N≤10^18
None of the digits in N is 0.If it is impossible to make a multiple of 3, print -1; otherwise, print the minimum number of digits that must be erased to make such a number.<b>Sample Input 1</b>
35
<b>Sample Output 1</b>
1
<b>Sample Input 2</b>
369
<b>Sample Output 2</b>
0
<b>Sample Input 3</b>
6227384
<b>Sample Output 3</b>
1
<b>Sample Input 4</b>
11
<b>Sample Output 4</b>
-1, I have written this Solution Code: #include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main() {
int64_t n;
scanf("%" SCNd64, &n);
int cnt[3] = { 0 };
while (n) {
cnt[n % 10 % 3]++;
n /= 10;
}
int cur = (cnt[1] + 2 * cnt[2]) % 3;
int k = cnt[0] + cnt[1] + cnt[2];
int res;
if (!cur) res = 0;
else if (cur == 1) {
if (cnt[1]) {
if (k == 1) res = -1;
else res = 1;
} else {
if (k == 2) res = -1;
else res = 2;
}
} else {
if (cnt[2]) {
if (k == 1) res = -1;
else res = 1;
} else {
if (k == 2) res = -1;
else res = 2;
}
}
printf("%d\n", res);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: static void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
System.out.print(x+4*j+" ");
}
System.out.println();
x+=6;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: def Pattern(N):
x=0
for i in range (0,N):
for j in range (0,N):
print(x+4*j,end=' ')
print()
x = x+6
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
printf("%d ",x+4*j);
}
printf("\n");
x+=6;
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
printf("%d ",x+4*j);
}
printf("\n");
x+=6;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a range from l to r, Your task is to find the sum of the second last digit of the prime numbers between that range.The first line of the input contains the T test cases.
Next T lines contains the range value l and r.
<b>Constraints</b>
1 <= T <= 100
1 <= l < r <= 1e5
Difference between l and r dosen't exceed 1e4.Print the required sum.Sample Input 1:
3
2 15
1 20
3 10
Sample Output 1:
2
4
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int m = sc.nextInt();
System.out.println(getPrimeSum(n,m));
}
}
public static int getPrimeSum(int n, int m){
int sum =0;
for(int i=n;i<=m;i++){
if(checkprime(i)){
int x=i;
sum += (i%100)/10;
}
}
return sum;
}
public static boolean checkprime(int n){
for(int i =2;i<=Math.sqrt(n);i++){
if(n%i==0){
return false;
}
}
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a range from l to r, Your task is to find the sum of the second last digit of the prime numbers between that range.The first line of the input contains the T test cases.
Next T lines contains the range value l and r.
<b>Constraints</b>
1 <= T <= 100
1 <= l < r <= 1e5
Difference between l and r dosen't exceed 1e4.Print the required sum.Sample Input 1:
3
2 15
1 20
3 10
Sample Output 1:
2
4
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
vector<bool> sieve(int n) {
vector<bool> is_prime(n + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= n; i++) {
if (is_prime[i]) {
for (int j = i * i; j <= n; j += i)
is_prime[j] = false;
}
}
return is_prime;
}
int main() {
vector<bool> prime = sieve(1e5 + 1);
int tt;
cin >> tt;
while (tt--) {
int l, r, sum = 0;
cin >> l >> r;
for (int i = l; i <= r; i++) {
if (prime[i]) {
sum += ((i % 100) / 10);
debug(i, (i % 100) / 10);
}
}
cout << sum << "\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: N = int(input())
if N > 5:
print("Greater than 5")
elif(N == 1):
print("one")
elif(N == 2):
print("two")
elif(N == 3):
print("three")
elif(N == 4):
print("four")
elif(N == 5):
print("five"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
String area = conditional(side);
System.out.println(area);
}static String conditional(int n){
if(n==1){return "one";}
else if(n==2){return "two";}
else if(n==3){return "three";}
else if(n==4){return "four";}
else if(n==5){return "five";}
else{
return "Greater than 5";}
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer value N. The task is to find how many numbers less than or equal to N have numbers of divisors exactly equal to 3.The first line contains integer T, denoting number of test cases. Then T test cases follow. The only line of each test case contains an integer N.
Constraints :
1 <= T <= 100000
1 <= N <= 1000000000For each testcase, in a new line, print the answer of each test case.Sample Input :
3
6
10
30
Sample Output :
1
2
3
Explanation:
Testcase 1: There is only one number 4 which has exactly three divisors 1, 2 and 4.
Testcase 2: 4 and 9 are the only two numbers less than or equal to 10 that have exactly three divisors.
Testcase 3: 4, 9, 25 are the only numbers less than or equal to 30 that have exactly three divisors., I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
fast();
vector<int> v;
int x=100000;
int a[x+5];
bool b[x+5];
for(int i=0;i<x+5;i++){
a[i]=0;
b[i]=false;
}
for(int i=2;i<x+5;i++){
if(b[i]==false){
for(int j=i+i;j<x+5;j+=i){
b[j]=true;
}}
}
int cnt=0;
for(int i=2;i<=x;i++){
if(b[i]==false){cnt++;}
a[i]=cnt;
}
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int p=sqrt(n);
out(a[p]);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code: def firstTwo(N):
while(N>99):
N=N//10
return (N%10)*10 + N//10
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code:
int firstTwo(int N){
while(N>99){
N/=10;
}
int ans = (N%10)*10 + N/10;
return ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code:
int firstTwo(int N){
while(N>99){
N/=10;
}
int ans = (N%10)*10 + N/10;
return ans;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N your task is to print its first two digits in reverse order. For eg:- If the given number is 123 then the output will be 21.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>firstTwo()</b> that takes integer N argument.
Constraints:-
111 <= N <= 9999
Note:- It is guaranteed that the given number will not contain any 0.Return the first two digits of the given number in reverse order.Sample Input:-
3423
Sample Output:-
43
Sample Input:-
1234
Sample Output:-
21, I have written this Solution Code: static int firstTwo(int N){
while(N>99){
N/=10;
}
int ans = (N%10)*10 + N/10;
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
int n = in.nextInt();
out.println(n*n);
out.flush();
}
static FastScanner in = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
static int oo = Integer.MAX_VALUE;
static long ooo = Long.MAX_VALUE;
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
void ini() throws FileNotFoundException {
br = new BufferedReader(new FileReader("input.txt"));
}
String next() {
while(!st.hasMoreTokens())
try { st = new StringTokenizer(br.readLine()); }
catch(IOException e) {}
return st.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] readArrayL(int n) {
long a[] = new long[n];
for(int i=0;i<n;i++) a[i] = nextLong();
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
double[] readArrayD(int n) {
double a[] = new double[n];
for(int i=0;i<n;i++) a[i] = nextDouble();
return a;
}
}
static final Random random = new Random();
static void ruffleSort(int[] a){
int n = a.length;
for(int i=0;i<n;i++){
int j = random.nextInt(n), temp = a[j];
a[j] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static void ruffleSortL(long[] a){
int n = a.length;
for(int i=0;i<n;i++){
int j = random.nextInt(n);
long temp = a[j];
a[j] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
cout<<(n*n)<<'\n';
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer N. Print the square of N.The input consists of a single integer N.
<b> Constraints: </b>
1 ≤ N ≤ 50Print a single integer – the value of N<sup>2</sup>.Sample Input 1:
5
Sample Output 1:
25, I have written this Solution Code: n=int(input())
print(n*n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four positive integers A, B, C, and D, can a rectangle have these integers as its sides?The input consists of a single line containing four space-separated integers A, B, C and D.
<b> Constraints: </b>
1 ≤ A, B, C, D ≤ 1000Output "Yes" if a rectangle is possible; otherwise, "No" (without quotes).Sample Input 1:
3 4 4 3
Sample Output 1:
Yes
Sample Explanation 1:
A rectangle is possible with 3, 4, 3, and 4 as sides in clockwise order.
Sample Input 2:
3 4 3 5
Sample Output 2:
No
Sample Explanation 2:
No rectangle can be made with these side lengths., I have written this Solution Code: #include<iostream>
using namespace std;
int main(){
int a,b,c,d;
cin >> a >> b >> c >> d;
if((a==c) &&(b==d)){
cout << "Yes\n";
}else if((a==b) && (c==d)){
cout << "Yes\n";
}else if((a==d) && (b==c)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] s = br.readLine().split(" ");
int N = Integer.parseInt(s[0]);
int M = Integer.parseInt(s[1]);
s = br.readLine().split(" ");
int[] prefix = new int[N];
int preSum = 0;
for(int i=0; i<N; i++){
int curr = Integer.parseInt(s[i]);
preSum += curr;
prefix[i] = preSum;
}
if(M>=prefix[N-1]){
System.out.println(prefix[N-1]);
return;
}
int maxSum = 0;
for(int i=0; i<N; i++){
for(int j=0; j<i; j++){
int curr = prefix[i]%M;
maxSum = Math.max(maxSum, curr);
curr = (prefix[i]-prefix[j])%M;
maxSum = Math.max(maxSum, curr);
if(maxSum==M-1){
break;
}
}
}
System.out.println(maxSum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: import bisect
def maximumSum(coll, m):
n = len(coll)
maxSum, prefixSum = 0, 0
sortedPrefixes = []
for endIndex in range(n):
prefixSum = (prefixSum + coll[endIndex]) % m
maxSum = max(maxSum, prefixSum)
startIndex = bisect.bisect_right(sortedPrefixes, prefixSum)
if startIndex < len(sortedPrefixes):
maxSum = max(maxSum, prefixSum - sortedPrefixes[startIndex] + m)
bisect.insort(sortedPrefixes, prefixSum)
return maxSum
a,b=map(int,input().split())
c=list(map(int,input().split()))
print(maximumSum(c,b)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
using namespace std;
// Return the maximum sum subarray mod m.
int maxSubarray(int arr[], int n, int m)
{
int x, prefix = 0, maxim = 0;
set<int> S;
S.insert(0);
// Traversing the array.
for (int i = 0; i < n; i++)
{
// Finding prefix sum.
prefix = (prefix + arr[i])%m;
// Finding maximum of prefix sum.
maxim = max(maxim, prefix);
// Finding iterator pointing to the first
// element that is not less than value
// "prefix + 1", i.e., greater than or
// equal to this value.
auto it = S.lower_bound(prefix+1);
if (it != S.end())
maxim = max(maxim, prefix - (*it) + m );
// Inserting prefix in the set.
S.insert(prefix);
}
return maxim;
}
// Driver Program
signed main()
{
int n,m;
cin>>n>>m;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout << maxSubarray(a, n, m) << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: def findMin(arr, low, high):
if high < low:
return arr[0]
if high == low:
return arr[low]
mid = int((low + high)/2)
if mid < high and arr[mid+1] < arr[mid]:
return arr[mid+1]
if mid > low and arr[mid] < arr[mid - 1]:
return arr[mid]
if arr[high] > arr[mid]:
return findMin(arr, low, mid-1)
return findMin(arr, mid+1, high)
T =int(input())
for i in range(T):
N = int(input())
arr = list(input().split())
for k in range(len(arr)):
arr[k] = int(arr[k])
print(findMin(arr,0,N-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
if(a[1] < a[n]){
cout << a[1] << endl;
continue;
}
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] >= a[1])
l = m;
else
h = m;
}
cout << a[h] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main (String[] args) {
//code
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int j=0;j<t;j++){
int al = s.nextInt();
int a[] = new int[al];
for(int i=0;i<al;i++){
a[i] = s.nextInt();
}
binSearchSmallest(a);
}
}
public static void binSearchSmallest(int a[]) {
int s=0;
int e = a.length - 1;
int mid = 0;
while(s<=e){
mid = (s+e)/2;
if(a[s]<a[e]){
System.out.println(a[s]);
return;
}
if(a[mid]>=a[s]){
s=mid+1;
}
else{
e=mid;
}
if(s == e){
System.out.println(a[s]);
return;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: // arr is the input array
function findMin(arr,n) {
// write code here
// do not console.log
// return the number
let min = arr[0]
for(let i=1;i<arr.length;i++){
if(min > arr[i]){
min = arr[i]
}
}
return min
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Modulo exponentiation is super awesome. But can you still think of a solution to the following problem?
Given three integers {a, b, c}, find the value of a<sup>b<sup>c</sup></sup> % 1000000007.
Here a<sup>b</sup> means a raised to the power b or pow(a, b). Expression evaluates to pow(a, pow(b, c)) % 1000000007.
(Read Euler's Theorem before solving this problem)The first input line has an integer t: the number of test cases.
After this, there are n lines, each containing three integers a, b and c.
Constraints
1≤ t ≤ 100
0 ≤ a, b, c ≤ 1000000000For each test case, output the value corresponding to the expression.Sample Input
3
3 7 1
15 2 2
3 4 5
Sample Output
2187
50625
763327764
Explaination:
In the first test, a = 3, b = 7, c = 1
b<sup>c</sup> = 7<sup>1</sup> = 7
a<sup>b<sup>c</sup></sup> = 3<sup>7</sup> = 2187, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int T=Integer.parseInt(br.readLine());
for(int k=0;k<T;k++)
{
String[] str= br.readLine().split(" ");
int a=Integer.parseInt(str[0]);
int b=Integer.parseInt(str[1]);
int c=Integer.parseInt(str[2]);
int M=1000000007;
System.out.print(superExponentation(a,superExponentation(b,c,M-1),M));
System.out.println();
}
}
public static long superExponentation(long a,long b,int m)
{
long res=1;
while(b>0)
{
if(b%2!=0)
res=(long)(res%m*a%m)%m;
a=((a%m)*(a%m))%m;
b=b>>1;
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Modulo exponentiation is super awesome. But can you still think of a solution to the following problem?
Given three integers {a, b, c}, find the value of a<sup>b<sup>c</sup></sup> % 1000000007.
Here a<sup>b</sup> means a raised to the power b or pow(a, b). Expression evaluates to pow(a, pow(b, c)) % 1000000007.
(Read Euler's Theorem before solving this problem)The first input line has an integer t: the number of test cases.
After this, there are n lines, each containing three integers a, b and c.
Constraints
1≤ t ≤ 100
0 ≤ a, b, c ≤ 1000000000For each test case, output the value corresponding to the expression.Sample Input
3
3 7 1
15 2 2
3 4 5
Sample Output
2187
50625
763327764
Explaination:
In the first test, a = 3, b = 7, c = 1
b<sup>c</sup> = 7<sup>1</sup> = 7
a<sup>b<sup>c</sup></sup> = 3<sup>7</sup> = 2187, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mem(a, b) memset(a, (b), sizeof(a))
#define fore(i,a) for(int i=0;i<a;i++)
#define fore1(i,j,a) for(int i=j;i<a;i++)
#define print(ar) for(int i=0;i<ar.size();i++)cout<<ar[i]<<" ";
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<long long> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
ll fastexp (ll a, ll b, ll n) {
ll res = 1;
while (b) {
if (b & 1) res = res*a%n;
a = a*a%n;
b >>= 1;
}
return res;
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main()
{
fast();
ll a,b,c;
int t, n, k;
cin >> t;
while(t--) {
cin >> a >>b >>c;
ll mod = 1e9+7;
ll k = fastexp(b,c,mod-1);
ll ans= fastexp(a,k,mod);
cout<<ans<<endl;
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array a of N integers. Count the number of triplets (i, j, k) such that 0 ≤ i < j < k < N
and |a[i] - a[j]| = |a[j] - a[k]|.First line contains N.
Next line contains N space separated integers.
<b>Constraints</b>
1 ≤ N ≤ 10<sup>3</sup>
1 ≤ a[i] ≤ 10<sup>9</sup>A single integer denoting the count of required triplets.Input:
5
5 1 4 3 5
Output:
5
Explanation:
(0, 1, 4) => |a[0] - a[1]| = |a[1] - a[4]| => |5-1| = |1-5| = 4
similarly (0, 2, 4), (0, 3, 4).
(0, 2, 3) => |a[0] - a[2]| = |a[2] - a[3]| => |5-4| = |4-3| = 1
(1, 3, 4) => |a[1] - a[3]| = |a[3] - a[4]| => |1-3| = |3-5| = 2, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(in.next());
}
long ans=0;
for(int j=1;j<(n-1);j++){
HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
for(int i=0;i<j;i++){
int x=Math.abs(a[i] - a[j]);
mp.put(x , mp.getOrDefault(x,0) + 1);
}
for(int k=j+1;k<n;k++){
int x=Math.abs(a[j] - a[k]);
ans += mp.getOrDefault(x,0);
}
}
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] implementMergeSort(int arr[], int start, int end)
{
if (start < end)
{
// Find the middle point
int mid = (start+end)/2;
// Sort first and second halves
implementMergeSort(arr, start, mid);
implementMergeSort(arr , mid+1, end);
// Merge the sorted halves
merge(arr, start, mid, end);
}
return arr;
}
public static void merge(int arr[], int start, int mid, int end)
{
// Find sizes of two subarrays to be merged
int n1 = mid - start + 1;
int n2 = end - mid;
/* Create temp arrays */
int L[] = new int [n1];
int R[] = new int [n2];
/*Copy data to temp arrays*/
for (int i=0; i<n1; ++i)
L[i] = arr[start + i];
for (int j=0; j<n2; ++j)
R[j] = arr[mid + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = start;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: for _ in range(int(input())):
n = int(input())
print(*sorted(list(map(int,input().split())))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n, m;
cin >> n >> m;
cout << __gcd(n, m);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: def hcf(a, b):
if(b == 0):
return a
else:
return hcf(b, a % b)
li= list(map(int,input().strip().split()))
print(hcf(li[0], li[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] sp = br.readLine().trim().split(" ");
long m = Long.parseLong(sp[0]);
long n = Long.parseLong(sp[1]);
System.out.println(GCDAns(m,n));
}
private static long GCDAns(long m,long n){
if(m==0)return n;
if(n==0)return m;
return GCDAns(n%m,m);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. In one move, you can swap any two elements of this array. Find out whether you can make this array strictly increasing or not after any (possibly 0) number of moves.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>6</sup>Print "YES", if you can make the array strictly increasing, else print "NO", without the quotes.Sample Input:
5
5 9 4 5 3
Sample Output:
NO
Explaination:
After any number of moves, you cannot make this array strictly increasing., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
vector<int> a(n);
for(auto &i : a) cin >> i;
sort(a.begin(), a.end());
for(int i = 1; i < n; i++){
if(a[i] == a[i - 1]){
cout << "NO";
return 0;
}
}
cout << "YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: function Charity(n,m) {
// write code here
// do no console.log the answer
// return the output using return keyword
const per = Math.floor(m / n)
return per > 1 ? per : -1
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: static int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: def Charity(N,M):
x = M//N
if x<=1:
return -1
return x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a charity which has N people. Penny wants to donate some of her clothes to the charity in such a way that all people receive equal clothes and each individual receives <b> more than 1 </b>. If she has M clothes with her what is the maximum number of clothes one individual can get?<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Charity()</b> that takes integers N, and M as arguments.
Constraints:-
1 <= M, N <= 1000Return the maximum number of clothes one individual can get if it is impossible to distribute clothes return -1.Sample Input
6 20
Sample Output
3
Sample Input
8 5
Sample Output
-1, I have written this Solution Code: int Charity(int n, int m){
int x= m/n;
if(x<=1){return -1;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Australia hasn’t lost a Test Match at the Gabba since 1988. India has to win this one, but unfortunately all of their players are injured.
You are given the following information:
1. A total of N players are available, all are injured initially.
2. You have M magic pills. Using X pills, you can make any one player fit for match.
3. Alternatively, you can exchange any player for Y magic pills.
Compute the maximum number of players you can make fit for the Gabba Test Match.
Note: "Exchanging" a player means that player will no longer be available and you will gain Y extra magic pills for the exchangeThe input contains a single line containing 4 integers N M X Y.
Constraints:-
0 ≤ N, M, X, Y ≤ 10^9The output should contain a single integer representing the maximum number of players you can make fit before the match.Sample Input:-
5 10 2 1
Sample Output:-
5
Explanation:-
You can make all players fit if you use all the pills.
Sample Input:-
3 10 4 2
Sample Output:-
2
Explanation:-
You can make 2 players fit using the initial candies. Otherwise, Exchanging one player would result in a total of 10+2=12 magic pills. This is enough to make 3 players fit, but you won't have 3 players remaining., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String values[] = br.readLine().trim().split(" ");
long N = Long.parseLong(values[0]);
long M = Long.parseLong(values[1]);
long X = Long.parseLong(values[2]);
long Y = Long.parseLong(values[3]);
if(M/X >= N){
System.out.print(N);
return;
}
else
{
long count = M/X;
M = M % X;
N = N - count;
if(((N-1)*Y+M)<X)
System.out.print(count);
else
System.out.print(count+ N - countSacrifice(1,N,M,X,Y));
}
}
public static long countSacrifice(long min,long max,long M,long X,long Y)
{
long N = max;
while(min<max)
{ long mid = min + (max-min)/2;
if((mid*Y + M)>=((N-mid)*X))
{
max = mid;
}
else
{
min = mid+1;
}
}
return min;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Australia hasn’t lost a Test Match at the Gabba since 1988. India has to win this one, but unfortunately all of their players are injured.
You are given the following information:
1. A total of N players are available, all are injured initially.
2. You have M magic pills. Using X pills, you can make any one player fit for match.
3. Alternatively, you can exchange any player for Y magic pills.
Compute the maximum number of players you can make fit for the Gabba Test Match.
Note: "Exchanging" a player means that player will no longer be available and you will gain Y extra magic pills for the exchangeThe input contains a single line containing 4 integers N M X Y.
Constraints:-
0 ≤ N, M, X, Y ≤ 10^9The output should contain a single integer representing the maximum number of players you can make fit before the match.Sample Input:-
5 10 2 1
Sample Output:-
5
Explanation:-
You can make all players fit if you use all the pills.
Sample Input:-
3 10 4 2
Sample Output:-
2
Explanation:-
You can make 2 players fit using the initial candies. Otherwise, Exchanging one player would result in a total of 10+2=12 magic pills. This is enough to make 3 players fit, but you won't have 3 players remaining., I have written this Solution Code: import math
N,M,X,Y = map(int,input().strip().split())
low = 0
high = N
def possible(mid,M,X,Y):
if M//X >= mid:
return True
elif (N-math.ceil(((X*mid)-M)/Y))>=mid:
return True
return False
while(low<=high):
mid = (high+low)>>1
if possible(mid,M,X,Y):
res = mid
low = mid+1
else:
high = mid-1
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Australia hasn’t lost a Test Match at the Gabba since 1988. India has to win this one, but unfortunately all of their players are injured.
You are given the following information:
1. A total of N players are available, all are injured initially.
2. You have M magic pills. Using X pills, you can make any one player fit for match.
3. Alternatively, you can exchange any player for Y magic pills.
Compute the maximum number of players you can make fit for the Gabba Test Match.
Note: "Exchanging" a player means that player will no longer be available and you will gain Y extra magic pills for the exchangeThe input contains a single line containing 4 integers N M X Y.
Constraints:-
0 ≤ N, M, X, Y ≤ 10^9The output should contain a single integer representing the maximum number of players you can make fit before the match.Sample Input:-
5 10 2 1
Sample Output:-
5
Explanation:-
You can make all players fit if you use all the pills.
Sample Input:-
3 10 4 2
Sample Output:-
2
Explanation:-
You can make 2 players fit using the initial candies. Otherwise, Exchanging one player would result in a total of 10+2=12 magic pills. This is enough to make 3 players fit, but you won't have 3 players remaining., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define LL long long
void solve() {
LL n, m, x, y;
cin >> n >> m >> x >> y;
LL l = 0, r = n + 1, mid;
while(l < r-1) {
mid = (l + r) / 2;
if(mid * x <= m + (n - mid) * y) l = mid;
else r = mid;
}
cout << l << '\n';
}
int main() {
ios::sync_with_stdio(0), cin.tie(0);
int tt = 1; //cin >> tt;
while(tt--) solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and a string S of length (N-1) containing only characters '0' and '1'. Find an array Arr which contains N integers, such that:
> All the values of the array are distinct
> Maximum value of the array can be N
> For i from 1 to N-1
if S[i] == '0'
Arr[i] > Arr[i+1]
if S[i] == '1'
Arr[i] < Arr[i+1]
> Arr is lexographically maximum possible
Note:- It can be proven that it will always be possible to create an array Arr satisfying above conditions.First line of input contains a single integer N.
Second line of input contains the string S of length N-1.
Constraints
2 <= N <= 100000
|S| = (N-1)
S contains characters '0' and '1' only.Print N space separated integers, denoting the array Arr.Sample Input 1
5
0001
Sample Output 1
5 4 3 1 2
Explanation
as S[1] == 0, A[1] > A[2]
as S[2] == 0, A[2] > A[3]
as S[3] == 0, A[3] > A[4]
as S[4] == 1, A[4] < A[5]
this is lexographically maximum array possible.
Sample Input 2
6
11100
Sample Output 2
3 4 5 6 2 1, I have written this Solution Code: k = int(input())
s = input()
left = 1
right = k
n = len(s)
c = 0
for i in range(n):
if s[i] == '0':
if c == 0:
print(right, end = " ")
right -= 1
else:
ans = ""
for j in range(c+1):
ans = str(right) + " " + ans
right -= 1
print(ans, end = "")
c = 0
else:
c += 1
ans = ""
for j in range(c+1):
ans = str(right) + " " + ans
right -= 1
print(ans, end = ""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |