Instruction
stringlengths 261
35k
| Response
stringclasses 1
value |
|---|---|
For this Question: Given an integer N, find the number of distinct arrays A, such that the sum of integers in the array is N and each integer in the array is greater than or equal to 3.
As the answer can be large, output it modulo 10<sup>9</sup>+7.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000Output a single integer, the number of arrays modulo 1000000007.Sample Input
8
Sample Output
4
Explanation:
Following are the possible arrays: [3, 5], [4, 4], [5, 3], [8].
Sample Input
2
Sample Output
0, 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().trim());
long[] arr = new long[n+1];
for(int i = 3; i <= n; i++)
arr[i] = -1;
if(n < 3) {
System.out.println(0);
} else {
long out = findRandomSequences(n, arr);
System.out.println(out);
}
}
public static long findRandomSequences(int n, long arr[]) {
if(n >= 3 && n < 6)
return 1;
if(arr[n] != -1)
return arr[n];
long count = 0;
for(int i = n-3; i >= 3; i--) {
count += (findRandomSequences(i, arr) % 1000000007);
}
arr[n] = (count+1) % 1000000007;
return arr[n];
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the number of distinct arrays A, such that the sum of integers in the array is N and each integer in the array is greater than or equal to 3.
As the answer can be large, output it modulo 10<sup>9</sup>+7.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000Output a single integer, the number of arrays modulo 1000000007.Sample Input
8
Sample Output
4
Explanation:
Following are the possible arrays: [3, 5], [4, 4], [5, 3], [8].
Sample Input
2
Sample Output
0, I have written this Solution Code: n = int(input())
memoArray = [0]*(n+1)
def findWays(n):
memoArray[3], memoArray[4], memoArray [5] = 1,1,1
for i in range(6,n+1):
memoArray[i] = (memoArray[i-1]+memoArray[i-3])%1000000007
return memoArray[n]
if n<3:
print(0)
elif n==3 or n == 4:
print(1)
else:
print(findWays(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the number of distinct arrays A, such that the sum of integers in the array is N and each integer in the array is greater than or equal to 3.
As the answer can be large, output it modulo 10<sup>9</sup>+7.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000Output a single integer, the number of arrays modulo 1000000007.Sample Input
8
Sample Output
4
Explanation:
Following are the possible arrays: [3, 5], [4, 4], [5, 3], [8].
Sample Input
2
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> dp(n+1, 0);
dp[0]=1;
For(i, 3, n+1){
For(j, 3, i+1){
dp[i] += dp[i-j];
}
dp[i]%=MOD;
}
cout<<dp[n];
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
function LeapYear(year){
// write code here
// return the output using return keyword
// do not use console.log here
if ((0 != year % 4) || ((0 == year % 100) && (0 != year % 400))) {
return 0;
} else {
return 1
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: year = int(input())
if year % 4 == 0 and not year % 100 == 0 or year % 400 == 0:
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 a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, 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();
int area = LeapYear(side);
if(area==1){
System.out.println("YES");}
else{
System.out.println("NO");}
}
static int LeapYear(int year){
if(year%400==0){return 1;}
if(year%100 != 0 && year%4==0){return 1;}
else {
return 0;}
}
}, In this Programming Language: Java, 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 went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: def sample(n):
if n<200:
print(200-n)
elif n<400:
print(400-n)
elif n<500:
print(500-n)
else:
div=n//100
if div*100==n:
print(0)
else:
print((div+1)*100-n)
n=int(input())
sample(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
if(n <= 200){
cout<<200-n;
return;
}
if(n <= 400){
cout<<400-n;
return;
}
int ans = (100-n%100)%100;
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, 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 ans=0;
if(n <= 200){
ans = 200-n;
}
else if(n <= 400){
ans=400-n;
}
else{
ans = (100-n%100)%100;
}
System.out.print(ans);
}
}
, 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 calculate the sum of all the odd integers present in the array.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
1 < = Arr[i] < = 10000Print the sum of all the odd integers present in the array.Sample Input:-
4
1 2 3 4
Sample Output:-
4
Sample Input:-
2 4 6 8
Sample Output:-
0, I have written this Solution Code: n = int(input())
num = input().split(" ")
sums=0
for i in range(0,n):
if int(num[i])%2 != 0:
sums = sums + int(num[i])
print(sums), 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 calculate the sum of all the odd integers present in the array.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
1 < = Arr[i] < = 10000Print the sum of all the odd integers present in the array.Sample Input:-
4
1 2 3 4
Sample Output:-
4
Sample Input:-
2 4 6 8
Sample Output:-
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
int sum=0;
for(int i=0;i<n;i++){
cin>>a;
if(a&1){sum+=a;}}
cout<<sum;
}
, 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 calculate the sum of all the odd integers present in the array.First line of input contains a single integer N. The next line contains the N space separated integers.
Constraints:-
1 < = N < = 1000
1 < = Arr[i] < = 10000Print the sum of all the odd integers present in the array.Sample Input:-
4
1 2 3 4
Sample Output:-
4
Sample Input:-
2 4 6 8
Sample Output:-
0, 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 sum=0;
for(int i=0;i<n;i++){
if(a[i]%2==1){
sum+=a[i];
}
}
System.out.print(sum);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono has gathered all the FRIENDS for a grand chapo event. For winning the chapo, she has asked the FRIENDS for the solution to a simple problem.
Our magical Tono has A apple candies and O orange candies. In one turn, she chooses an apple or an orange candy with equal probability, and turns it into the other type. Now, she asks the FRIENDS the expected number of turns she will follow this process until either the number of apple or the orange candies become 0.The first and the only line of input contains two integers A and O.
Constraints
1 <= A, O <= 10<sup>9</sup>Output the single line corresponding to the expected number of turns Tono needs to complete the process.
The answer is of the form p/q. Print it as (p*r)%1000000007 where r is modulo inverse of q with respect to 1000000007.Sample Input
1 2
Sample Output
2
Sample Input
0 5
Sample Output
0, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
public class Main {
static long mod = 1000000007;
public static void main(String[] args) throws IOException {
FastScanner s=new FastScanner();
int t1 =1;
while(t1-->0){
long a = s.nextLong();
long o = s.nextLong();
if(a==0 || o==0) System.out.println(0);
else System.out.println(a*(long)o%mod);
}
}
static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
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());
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono has gathered all the FRIENDS for a grand chapo event. For winning the chapo, she has asked the FRIENDS for the solution to a simple problem.
Our magical Tono has A apple candies and O orange candies. In one turn, she chooses an apple or an orange candy with equal probability, and turns it into the other type. Now, she asks the FRIENDS the expected number of turns she will follow this process until either the number of apple or the orange candies become 0.The first and the only line of input contains two integers A and O.
Constraints
1 <= A, O <= 10<sup>9</sup>Output the single line corresponding to the expected number of turns Tono needs to complete the process.
The answer is of the form p/q. Print it as (p*r)%1000000007 where r is modulo inverse of q with respect to 1000000007.Sample Input
1 2
Sample Output
2
Sample Input
0 5
Sample Output
0, I have written this Solution Code: a,b =map(int, input().split())
print((a*b)%1000000007), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono has gathered all the FRIENDS for a grand chapo event. For winning the chapo, she has asked the FRIENDS for the solution to a simple problem.
Our magical Tono has A apple candies and O orange candies. In one turn, she chooses an apple or an orange candy with equal probability, and turns it into the other type. Now, she asks the FRIENDS the expected number of turns she will follow this process until either the number of apple or the orange candies become 0.The first and the only line of input contains two integers A and O.
Constraints
1 <= A, O <= 10<sup>9</sup>Output the single line corresponding to the expected number of turns Tono needs to complete the process.
The answer is of the form p/q. Print it as (p*r)%1000000007 where r is modulo inverse of q with respect to 1000000007.Sample Input
1 2
Sample Output
2
Sample Input
0 5
Sample Output
0, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int mo=1000000007;
int a,b;
cin>>a>>b;
cout<<(a*b)%mo;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, 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 <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.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>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "True" is the number is palindrome otherwise "False".Sample Input:
5
Sample Output:
True
Sample Input:
121
Sample Output:
True, I have written this Solution Code: static void isPalindrome(int N)
{
int digit = 0, sum = 0, temp = N;
while(N > 0)
{
digit = N %10;
sum = sum*10 + digit;
N = N/10;
}
if(sum == temp)
System.out.println("True");
else System.out.println("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 <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.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>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "True" is the number is palindrome otherwise "False".Sample Input:
5
Sample Output:
True
Sample Input:
121
Sample Output:
True, I have written this Solution Code: void isPalindrome(int N){
int digit=0, sum=0, temp = N;
while(N > 0)
{
digit = N%10;
sum = sum*10 + digit;
N = N/10;
}
if(sum == temp)
cout << "True";
else cout << "False";
}
, In this Programming Language: C++, 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 <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.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>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "True" is the number is palindrome otherwise "False".Sample Input:
5
Sample Output:
True
Sample Input:
121
Sample Output:
True, I have written this Solution Code: def isPalindrome(N):
res = str(N) == str(N)[::-1]
return res, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement <code>createUserObj</code> which takes two arguments, email and password
and the funtion returns and object with key email and value as email argument and key password
and value as password.Function will take two arguments.Function will return object with keys email and passwordconst obj = createUserObj("akshat. sethi@newtonschool. co", "123456")
console. log(obj) // prints {email:"akshat. sethi@newtonschool. co", password:"123456"}, I have written this Solution Code: function createUserObj(email,password){
return {email,password}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she 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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code:
int minimumMarks(int X){
if(X>5){
return 10-2*(10-X);
}
return 10-2*X;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she 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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code: def minimumMarks(X):
if X>5:
return 10-2*(10-X)
return 10-2*(X)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she 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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code: static int minimumMarks(int X){
if(X>5){
return 10-2*(10-X);
}
return 10-2*X;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she 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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code:
int minimumMarks(int X){
if(X>5){
return 10-2*(10-X);
}
return 10-2*X;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code:
import sys
from collections import defaultdict
from heapq import heappush, heappop
n, m = map(int, input().split())
d = defaultdict(list)
dist = [sys.maxsize]*n
dist[0] = 0
for _ in range(m):
start, dest, wt = map(int, input().split())
d[start-1].append((dest-1, wt))
d[dest-1].append((start-1, wt))
heap = [(0, 0, 0)]
while heap:
count, cost, u= heappop(heap)
for vertex, weight in d[u]:
if dist[vertex] > cost + weight*(count+1):
dist[vertex] = cost + weight*(count+1)
heappush(heap, (count+1, dist[vertex], vertex))
for d in dist:
if d == sys.maxsize:
print(-1)
else:
print(d), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF =1e18;
vector<tuple<int, int, int>> adj;
void solve()
{
int n, m;
cin>>n>>m;
// assert(1<=n && n<=3000);
// assert(0<=m && m<=10000);
adj.resize(n);
for(int i = 0;i<m;i++)
{
int x, y, w;
cin>>x>>y>>w;
x--;
y--;
// assert(0<=x && x<n);
// assert(0<=y && y<n);
// assert(1<=w && w<=1e9);
adj.push_back({x, y, w});
adj.push_back({y, x, w});
}
vector<int> dp_old(n, INF);
vector<int> dp_new(n, INF);
dp_old[0] = 0;
for(int i = 1;i<=n;i++)
{
fill(dp_new.begin(), dp_new.end(), INF);
for(auto [x, y,w]:adj)
{
dp_new[y]= min({dp_new[y], dp_old[x] + i * w});
}
for(int j = 0;j<n;j++)
dp_new[j] = min(dp_new[j], dp_old[j]);
swap(dp_new, dp_old);
}
for(int i = 0;i<n;i++)
{
if(dp_old[i] == INF)
dp_old[i] = -1;
cout<<dp_old[i]<<"\n";
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;import static java.lang.Math.*;import static java.
util.Map.*;import static java.util.Arrays.*;import static java.util.Collections.*;
import static java.lang.System.*;
public class Main
{
public void tq()throws Exception
{
st=new StringTokenizer(bq.readLine());
int tq=1;
sb=new StringBuilder(2000000);
o:
while(tq-->0)
{
int n=i();
int m=i();
LinkedList<int[]> l[]=new LinkedList[n];
for(int x=0;x<n;x++)l[x]=new LinkedList<>();
for(int x=0;x<m;x++)
{
int a=i()-1;
int b=i()-1;
int c=i();
l[a].add(new int[]{b,c});
l[b].add(new int[]{a,c});
}
long d[]=new long[n];
for(int x=0;x<n;x++)d[x]=maxl;
d[0]=0l;
PriorityQueue<long[]> p=new PriorityQueue<>(5000,(a,b)->a[2]-b[2]<1l?-1:1);
p.add(new long[]{0l,0,0});
while(p.size()>0)
{
long r[]=p.poll();
long di=r[0];
int no=(int)r[1];
long mu=r[2];
for(int e[]:l[no])
{
int node=e[0];
int w=e[1];
long de=di+w*(mu+1);
if(d[node]>de)
{
d[node]=de;
p.add(new long[]{de,node,mu+1});
}
}
}
for(long x:d)
{
sl(x==maxl?-1:x);
}
}
p(sb);
}
int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE, minl=Long.
MIN_VALUE;BufferedReader bq=new BufferedReader(new InputStreamReader(in));StringTokenizer st;
StringBuilder sb;public static void main(String[] a)throws Exception{new Main().tq();}int[] so(int ar[])
{Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;
x++)ar[x]=r[x];return ar;}long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)
r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
char[] so(char ar[]) {Character
r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)
ar[x]=r[x];return ar;}void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.
append(s);}void s(char s){sb.append(s);}void s(double s){sb.append(s);}void ss(){sb.append(' ');}void sl
(String s){sb.append(s);sb.append("\n");}void sl(int s){sb.append(s);sb.append("\n");}void sl(long s){sb
.append(s);sb.append("\n");}void sl(char s) {sb.append(s);sb.append("\n");}void sl(double s){sb.append(s)
;sb.append("\n");}void sl(){sb.append("\n");}int l(int v){return 31-Integer.numberOfLeadingZeros(v);}
long l(long v){return 63-Long.numberOfLeadingZeros(v);}int sq(int a){return (int)sqrt(a);}long sq(long a)
{return (long)sqrt(a);}long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}int gcd(int a,int b)
{while(b>0){int c=a%b;a=b;b=c;}return a;}boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=
s.charAt(j--))return false;return true;}boolean[] si(int n) {boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]
=true;for(int x=4;x<=n;x+=2)bo[x]=true;for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;
y+=vv)bo[y]=true;}}return bo;}long mul(long a,long b,long m) {long r=1l;a%=m;while(b>0){if((b&1)==1)
r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}int i()throws IOException{if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());return Integer.parseInt(st.nextToken());}long l()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Long.parseLong(st.nextToken());}String
s()throws IOException {if (!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Double.
parseDouble(st.nextToken());}void p(Object p){out.print(p);}void p(String p){out.print(p);}void p(int p)
{out.print(p);}void p(double p){out.print(p);}void p(long p){out.print(p);}void p(char p){out.print(p);}void
p(boolean p){out.print(p);}void pl(Object p){out.println(p);}void pl(String p){out.println(p);}void pl(int p)
{out.println(p);}void pl(char p){out.println(p);}void pl(double p){out.println(p);}void pl(long p){out.
println(p);}void pl(boolean p)
{out.println(p);}void pl(){out.println();}void s(int a[]){for(int e:a)
{sb.append(e);sb.append(' ');}sb.append("\n");}
void s(long a[])
{for(long e:a){sb.append(e);sb.append(' ')
;}sb.append("\n");}void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append
("\n");}}
void s(char a[])
{for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}void s(char ar[][])
{for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}int[] ari(int n)throws
IOException {int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());for(int x=0;
x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}int[][] ari(int n,int m)throws
IOException {int ar[][]=new int[n][m];for(int x=0;x<n;x++){if (!st.hasMoreTokens())st=new StringTokenizer
(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}long[] arl
(int n)throws IOException {long ar[]=new long[n];if(!st.hasMoreTokens()) st=new StringTokenizer(bq.readLine())
;for(int x=0;x<n;x++)ar[x]=Long.parseLong(st.nextToken());return ar;}long[][] arl(int n,int m)throws
IOException {long ar[][]=new long[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens()) st=new
StringTokenizer(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
String[] ars(int n)throws IOException {String ar[] =new String[n];if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken();return ar;}double[] ard
(int n)throws IOException {double ar[] =new double[n];if(!st.hasMoreTokens())st=new StringTokenizer
(bq.readLine());for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}double[][] ard
(int n,int m)throws IOException{double ar[][]=new double[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens())
st=new StringTokenizer(bq.readLine());for(int y=0;y<m;y++) ar[x][y]=Double.parseDouble(st.nextToken());}
return ar;}char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}char[][]
arc(int n,int m)throws IOException {char ar[][]=new char[n][m];for(int x=0;x<n;x++){String s=bq.readLine();
for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}void p(int ar[])
{StringBuilder sb=new StringBuilder
(2*ar.length);for(int a:ar){sb.append(a);sb.append(' ');}out.println(sb);}void p(int ar[][])
{StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(int a[]:ar){for(int aa:a){sb.append(aa);
sb.append(' ');}sb.append("\n");}p(sb);}void p(long ar[]){StringBuilder sb=new StringBuilder
(2*ar.length);for(long a:ar){ sb.append(a);sb.append(' ');}out.println(sb);}
void p(long ar[][])
{StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(long a[]:ar){for(long aa:a){sb.append(aa);
sb.append(' ');}sb.append("\n");}p(sb);}void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;
StringBuilder sb=new StringBuilder(c);for(String a:ar){sb.append(a);sb.append(' ');}out.println(sb);}
void p(double ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(double a:ar){sb.append(a);
sb.append(' ');}out.println(sb);}void p
(double ar[][]){StringBuilder sb=new StringBuilder(2*
ar.length*ar[0].length);for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n")
;}p(sb);}void p(char ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(char aa:ar){sb.append(aa);
sb.append(' ');}out.println(sb);}void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0]
.length);for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: def lenOfLongSubarr(arr, N, K):
mydict = dict()
sum = 0
maxLen = 0
for i in range(N):
sum += arr[i]
if (sum == K):
maxLen = i + 1
elif (sum - K) in mydict:
maxLen = max(maxLen, i - mydict[sum - K])
if sum not in mydict:
mydict[sum] = i
return maxLen
if __name__ == '__main__':
T = int(input())
#N,K=list(map(int,input().split()))
for i in range(T):
N,k= [int(N)for N in input("").split()]
arr=list(map(int,input().split()))
N = len(arr)
print(lenOfLongSubarr(arr, N, k)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
unordered_map<long long,int> um;
int n,k;
cin>>n>>k;
long arr[n];
int maxLen=0;
for(int i=0;i<n;i++){cin>>arr[i];}
long long sum=0;
for(int i=0;i<n;i++){
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
cout<<maxLen<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
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){
if(t%10==0){
System.gc();
}
int arrsize=sc.nextInt();
int k=sc.nextInt();
int[] arr=new int[arrsize];
for(int i=0;i<arrsize;i++){
arr[i]=sc.nextInt();
}
int subsize=0;
int sum=0;
HashMap<Integer, Integer> hash=new HashMap<>();
for(int i=0;i<arrsize;i++){
sum+=arr[i];
if(sum==k){
subsize=i+1;
}
if(!hash.containsKey(sum)){
hash.put(sum,i);
}
if(hash.containsKey(sum-k)){
if(subsize<(i-hash.get(sum-k))){
subsize=i-hash.get(sum-k);
}
}
}
System.out.println(subsize);
}
}
}, In this Programming Language: Java, 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 integer array nums of size n, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i].Implement the given function.
Constraints:
1 <= n <= 10^5
1 <= nums[i] <= 10^9Return the count array.Input:
4
5 2 6 1
Output:
2 1 1 0, I have written this Solution Code: class Node:
def __init__(self,val):
self.val = val
self.left = None
self.right = None
self.elecount = 1
self.lcount = 0
class Tree:
def __init__(self,root):
self.root = root
def insert(self,node):
curr = self.root
cnt = 0
while curr!=None:
prev = curr
if node.val>curr.val:
cnt += (curr.elecount+curr.lcount)
curr=curr.right
elif node.val<curr.val:
curr.lcount+=1
curr=curr.left
else:
prev=curr
prev.elecount+=1
break
if prev.val>node.val:
prev.left = node
elif prev.val<node.val:
prev.right = node
else:
return cnt+prev.lcount
return cnt
def constructArray(arr,n):
t = Tree(Node(arr[-1]))
ans = [0]
for i in range(n-2,-1,-1):
ans.append(t.insert(Node(arr[i])))
return reversed(ans)
n=int(input())
arr = list(map(int,input().split()))
print(" ".join(list(map(str,constructArray(arr,n))))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array nums of size n, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i].Implement the given function.
Constraints:
1 <= n <= 10^5
1 <= nums[i] <= 10^9Return the count array.Input:
4
5 2 6 1
Output:
2 1 1 0, I have written this Solution Code: class Solution{
int[] count;
public List<Integer> countSmaller(int[] nums) {
List<Integer> res = new ArrayList<Integer>();
count = new int[nums.length];
int[] indexes = new int[nums.length];
for(int i = 0; i < nums.length; i++){
indexes[i] = i;
}
mergesort(nums, indexes, 0, nums.length - 1);
for(int i = 0; i < count.length; i++){
res.add(count[i]);
}
return res;
}
private void mergesort(int[] nums, int[] indexes, int start, int end){
if(end <= start){
return;
}
int mid = (start + end) / 2;
mergesort(nums, indexes, start, mid);
mergesort(nums, indexes, mid + 1, end);
merge(nums, indexes, start, end);
}
private void merge(int[] nums, int[] indexes, int start, int end){
int mid = (start + end) / 2;
int left_index = start;
int right_index = mid+1;
int rightcount = 0;
int[] new_indexes = new int[end - start + 1];
int sort_index = 0;
while(left_index <= mid && right_index <= end){
if(nums[indexes[right_index]] < nums[indexes[left_index]]){
new_indexes[sort_index] = indexes[right_index];
rightcount++;
right_index++;
}else{
new_indexes[sort_index] = indexes[left_index];
count[indexes[left_index]] += rightcount;
left_index++;
}
sort_index++;
}
while(left_index <= mid){
new_indexes[sort_index] = indexes[left_index];
count[indexes[left_index]] += rightcount;
left_index++;
sort_index++;
}
while(right_index <= end){
new_indexes[sort_index++] = indexes[right_index++];
}
for(int i = start; i <= end; i++){
indexes[i] = new_indexes[i - start];
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader inputMachine = new BufferedReader(new InputStreamReader(System.in));
int length = Integer.parseInt(inputMachine.readLine());
String[] arrText = inputMachine.readLine().trim().split(" ");
int[] nums = new int[length];
for (int i = 0; i < nums.length; i++) {
nums[i] = Integer.parseInt(arrText[i]);
}
Arrays.sort(nums);
int i = 0;
int j = length - 1;
long sum = 0;
while (i < j) {
sum += nums[j] - nums[i];
i++;
j--;
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
arr.sort()
s = 0
for i in range(n//2):
s += arr[n-i-1]-arr[i]
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
sort(a,a+n);
int ans=0;
for(int i=0;i<n/2;++i){
ans+=abs(a[i]-a[n-i-1]);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: int MakeOne(int N){
int cnt=0;
int a =N;
while(a!=1){
if(a&1){a++;}
else{a/=2;}
cnt++;
}
return cnt;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: static int MakeOne(int N){
int cnt=0;
int a =N;
while(a!=1){
if(a%2==1){a++;}
else{a/=2;}
cnt++;
}
return cnt;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: int MakeOne(int N){
int cnt=0;
int a =N;
while(a!=1){
if(a&1){a++;}
else{a/=2;}
cnt++;
}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you have to perform only two types of operations on the number:-
1) If N is odd increase it by 1
2) If N is even divide it by 2
Your task is to find the number of operations it takes to make N equal to 1.<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>MakeOne()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations it takes to make the number N equal to 1.Sample Input:-
7
Sample Output:-
4
Explanation:-
7 - > 8 - > 4 - > 2 - > 1
Sample Input:-
3
Sample Output:-
3, I have written this Solution Code: def MakeOne(N):
cnt=0
while N!=1:
if N%2==1:
N=N+1
else:
N=N//2
cnt=cnt+1
return cnt
, 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: 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: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: static int MagicKnight(int N, int T){
while(T-->0){
N/=2;
N+=2;
if(N==3 || N==4){return N;}
}
return N;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: long MagicKnight(long N, long T){
while(T--){
N/=2;
N+=2;
if(N==3 || N==4){return N;}
}
return N;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: def MagicKnight(N,T):
while T > 0:
N = N//2
N = N+2
if N == 3 or N ==4:
return N
T = T - 1
return N
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Asta wants to become a magic knight. In his journey to becoming a magic knight, he stuck on a problem and asks for your help. Problem description:-
Given a number N which at the end of each second gets halved(take floor value) and then increases by 2. Your task is to calculate the number N at the end of T second.<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>MagicKnight()</b> that takes integers N and T as parameters.
Constraints:-
3 <= N <= 1000000000
1 <= T <= 1000000000Return the number N at the end of T seconds.Sample Input:-
7 2
Sample Output:-
3
Explanation:-
After 1 sec :- N = 7/2 + 2 = 5
After 2 sec :- N = 5/2 + 2 = 4
Sample Input:-
10 1
Sample Output:-
7, I have written this Solution Code: long MagicKnight(long N, long T){
while(T--){
N/=2;
N+=2;
if(N==3 || N==4){return N;}
}
return N;
}, 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 are given an array A (distinct integers) of size N, and you are also given a sum. You need to find if two numbers in A exists that have sum equal to the given sum.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line contains N denoting the size of the array A and target sum. The second line contains N elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= sum <= 10^5
1 <= A[i] <= 10^4For each testcase, in a new line, print "1"(without quotes) if any pair found, othwerwise print "0"(without quotes) if not found.Sample Input
2
10 14
1 2 3 4 5 6 7 8 9 10
2 10
2 5
Sample Output
1
0
Explanation:
Testcase 1: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and sum = 14. There is a pair {4, 10} with sum 14.
Testcase 2: arr[] = {2, 5} and sum = 10. There is no pair with sum 10., I have written this Solution Code: import numpy as np
from collections import defaultdict
t=int(input())
def solve():
d=defaultdict(int)
n,s=input().strip().split()
s=int(s)
a=np.array([input().strip().split()],int).flatten()
for i in a:
d[i]+=1
c=0
for i in a:
if(d[s-i]>0 and (s-i)!=i):
c=1
break
print(c)
while(t>0):
solve()
t-=1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A (distinct integers) of size N, and you are also given a sum. You need to find if two numbers in A exists that have sum equal to the given sum.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line contains N denoting the size of the array A and target sum. The second line contains N elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= sum <= 10^5
1 <= A[i] <= 10^4For each testcase, in a new line, print "1"(without quotes) if any pair found, othwerwise print "0"(without quotes) if not found.Sample Input
2
10 14
1 2 3 4 5 6 7 8 9 10
2 10
2 5
Sample Output
1
0
Explanation:
Testcase 1: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and sum = 14. There is a pair {4, 10} with sum 14.
Testcase 2: arr[] = {2, 5} and sum = 10. There is no pair with sum 10., 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 pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{ t--;
int ch=0;
int n,sum;
cin>>n>>sum;
int A[n];
set<int> ss;
for(int i=0;i<n;i++)
{
cin>>A[i];
if(ss.find(sum-A[i])!=ss.end()) ch=1;
ss.insert(A[i]);
}
cout<<ch<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A (distinct integers) of size N, and you are also given a sum. You need to find if two numbers in A exists that have sum equal to the given sum.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line contains N denoting the size of the array A and target sum. The second line contains N elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= sum <= 10^5
1 <= A[i] <= 10^4For each testcase, in a new line, print "1"(without quotes) if any pair found, othwerwise print "0"(without quotes) if not found.Sample Input
2
10 14
1 2 3 4 5 6 7 8 9 10
2 10
2 5
Sample Output
1
0
Explanation:
Testcase 1: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and sum = 14. There is a pair {4, 10} with sum 14.
Testcase 2: arr[] = {2, 5} and sum = 10. There is no pair with sum 10., I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int testCases = sc.nextInt();
for(int i = 1; i <= testCases; i++)
{
int arrSize = sc.nextInt();
int sum = sc.nextInt();
int arr[] = new int[arrSize];
for(int j = 0; j < arrSize; j++)
arr[j] = sc.nextInt();
System.out.println(pairFound(arr, arrSize, sum));
}
}
static int pairFound(int arr[], int arrSize, int sum)
{
HashSet<Integer> hSet = new HashSet<>();
for(int i = 0; i < arrSize; i++)
{
if(hSet.contains(sum-arr[i]) == true)
return 1;
hSet.add(arr[i]);
}
return 0;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
st = new StringTokenizer(br.readLine());
String s = st.nextToken();
int curzeroes = 0;
StringBuilder sb = new StringBuilder();
int len = s.length();
for(int i = 0;i<len;i++){
if(s.charAt(i) == '1'){
if(curzeroes == 0){
sb.append("1");
}
else{
curzeroes--;
}
}
else{
curzeroes++;
}
}
for(int i = 0;i<curzeroes;i++){
sb.append("0");
}
if(sb.length() == 0 && curzeroes == 0){
bw.write("-1\n");
}
else{
bw.write(sb.toString()+"\n");
}
bw.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: arr = input()
c = 0
res = ""
n =len(arr)
for i in range(n):
if arr[i]=='0':
c+=1
else:
if c==0:
res+='1'
else:
c-=1
for i in range(c):
res+='0'
if len(res)==0:
print(-1)
else:
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
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;
string s; cin >> s;
stack<int> st;
for(int i = 0; i < (int)s.length(); i++){
if(!st.empty() && s[st.top()] == '0' && s[i] == '1'){
st.pop();
}
else
st.push(i);
}
string res = "";
while(!st.empty()){
res += s[st.top()];
st.pop();
}
reverse(res.begin(), res.end());
if(res == "") res = "-1";
cout << res;
return 0;
}, 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, find the count of pairs of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and the <b>mean</b> of elements at these indices is at least <b>K</b>.
More formally, you have to find the number of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and (A<sub>i</sub> + A<sub>j</sub>)/2 >= KFirst line of the input contains two integers N and K.
The second line contains N space seperated integers A<sub>i</sub>
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>9</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the number of pairs satisfying the above condition in array A.Sample Input:
5 6
4 7 8 2 5
Sample Output:
4
Explaination:
The following pairs of indices satisfy the condition (1-based indexing)
(1, 3) -> (4 + 8)/2 = 6
(2, 3) -> (7 + 8)/2 = 7.5
(2, 5) -> (7 + 5)/2 = 6
(3, 5) -> (8 + 5)/2 = 6.5
There are no more pairs of indices that satisfy the above condition., 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 long ll;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve() {
int n, k;
cin >> n >> k;
assert(1 <= n <= 1e5);
assert(1 <= k <= 1e9);
vector<int> a(n);
for (auto &i : a) cin >> i, assert(1 <= i <= 1e9);
sort(all(a));
int ans = 0;
int req_sum = 2 * k;
for (int i = 0; i < n; i++) {
int x = req_sum - a[i];
x = max(x, (int)0);
int l = i + 1, r = n - 1, ind = n;
while (l <= r) {
int m = (l + r) / 2;
if (a[m] >= x) {
r = m - 1;
ind = m;
} else
l = m + 1;
}
ans += n - ind;
}
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: Given an array A of size N, find the count of pairs of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and the <b>mean</b> of elements at these indices is at least <b>K</b>.
More formally, you have to find the number of indices <b>i</b> and <b>j</b> such that <b>1 <= i < j <= N</b> and (A<sub>i</sub> + A<sub>j</sub>)/2 >= KFirst line of the input contains two integers N and K.
The second line contains N space seperated integers A<sub>i</sub>
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>9</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the number of pairs satisfying the above condition in array A.Sample Input:
5 6
4 7 8 2 5
Sample Output:
4
Explaination:
The following pairs of indices satisfy the condition (1-based indexing)
(1, 3) -> (4 + 8)/2 = 6
(2, 3) -> (7 + 8)/2 = 7.5
(2, 5) -> (7 + 5)/2 = 6
(3, 5) -> (8 + 5)/2 = 6.5
There are no more pairs of indices that satisfy the above condition., 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 nk=br.readLine();
String nkarr[]=nk.split(" ");
int n =Integer.parseInt(nkarr[0]);
int k =Integer.parseInt(nkarr[1]);
String ss = br.readLine();
String sst[]=ss.split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(sst[i]);
}
long val = findMean(arr,n,k);
System.out.println(val);
}
public static long findMean(int []arr,int n,int k){
Arrays.sort(arr);
int low =0;
int high=arr.length-1;
long ans=0;
while(low<high){
long mean =(arr[low]+arr[high])/2;
if(mean>=k){
ans+=(high-low);
high--;
}else{
low++;
}
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>round</code>, which should take a number which can be a float(decimal)
and return its result as an integer rounded of (Use JS In built functions)Function will take a float as input (can be negative or positive)Function will return a rounded off numberconsole. log(round(1.112)) // prints 1
console. log(round(1.9)) // prints 2
console. log(round(-0.66)) // prints -1, I have written this Solution Code: function round(num){
// write code here
// return the output , do not use console.log here
return Math.round(num)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement the function <code>round</code>, which should take a number which can be a float(decimal)
and return its result as an integer rounded of (Use JS In built functions)Function will take a float as input (can be negative or positive)Function will return a rounded off numberconsole. log(round(1.112)) // prints 1
console. log(round(1.9)) // prints 2
console. log(round(-0.66)) // prints -1, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.math.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
double n=sc.nextDouble();
System.out.println(Math.round(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a Singly linked list and an integer K. Your task is to insert the integer K at the head of the given linked list<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>addElement()</b> that takes head node and the element K as a parameter.
Constraints:
1 <=N<= 1000
1 <=K, value<= 1000Return the head of the modified linked listSample Input:-
5 2
1 2 3 4 5
Sample Output:
2 1 2 3 4 5
, I have written this Solution Code: public static Node addElement(Node head,int k) {
Node temp =new Node(k);
temp.next=head;
return temp;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: static int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
def RotationPolicy(A, B):
cnt=0
for i in range (A,B+1):
if(i-1)%2!=0 and (i-1)%3!=0:
cnt=cnt+1
return cnt
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: function RotationPolicy(a, b) {
// write code here
// do no console.log the answer
// return the output using return keyword
let count = 0
for (let i = a; i <= b; i++) {
if((i-1)%2 !== 0 && (i-1)%3 !==0){
count++
}
}
return count
}
, In this Programming Language: JavaScript, 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.
<b>A number is said to 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
Sample Output:
true
Explanation:
1<sup>3</sup> = 1
Sample Input:
147
Sample Output:
false
Explanation:
1<sup>3</sup> + 4<sup>3</sup> + 7<sup>3</sup> != 147, I have written this Solution Code: def isArmstrong(N):
w=N
ans=0
while w>0:
x=w%10
ans=ans+x**3
w=int(w//10)
res=ans==N
return res
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import numpy as np
from collections import defaultdict
n=int(input())
a=np.array([input().strip().split()],int).flatten()
d=defaultdict(int)
for i in a:
d[i]+=1
d=sorted(d.items())
for i in d:
if(i[1]>1):
print(i[0],end=" ")
print(i[1])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int noOfElements = Integer.parseInt(in.readLine());
String [] elem = in.readLine().trim().split(" ");
int [] elements = new int [noOfElements];
for(int i=0 ; i< noOfElements; i++){
elements[i] = Integer.parseInt(elem[i]);
}
Arrays.sort(elements);
int count =0;
for(int i = 0 ; i < noOfElements-1 ; i++){
if(elements[i] == elements[i+1]){
count ++;
}
else if(count != 0){
System.out.print(elements[i]+" "+(count+1));
count =0;
System.out.println();
}
}
if(count != 0)
System.out.print(elements[noOfElements-1]+" "+(count+1));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, your task is to find the count of repeated elements. Print the repeated elements in ascending order along with their frequency.
Have a look at the example for more understanding.The first line of input contains a single integer N, the next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000For each duplicate element in sorted order in a new line, First, print the duplicate element and then print its number of occurence space- separated.
Note:- It is guaranteed that at least one duplicate element will exist in the given array.Sample Input:-
5
3 2 1 1 2
Sample Output:-
1 2
2 2
Sample Input:-
5
1 1 1 1 5
Sample Output:-
1 4
Explaination:
test 1: Only 1 and 2 are repeated. Both are repeated twice. So, we print:
1 -> frequency of 1
2 -> frequency of 2
1 is printed before 2 as it is smaller than 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n];
map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
m[a[i]]++;
}
for(auto it = m.begin();it!=m.end();it++){
if(it->second>1){
cout<<it->first<<" "<<it->second<<'\n';
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves pattern, so this time she wishes to draw a pattern as:-
*****
****
***
**
*
Since Sara does not know how to code, help her to draw this pattern.<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>Triangle()</b> that takes no parameters.Print the pattern as shown in the example.Sample Output:-
*****
****
***
**
*, I have written this Solution Code: def Triangle():
for i in range(5):
for j in range(5-i):
print("*", end='')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves pattern, so this time she wishes to draw a pattern as:-
*****
****
***
**
*
Since Sara does not know how to code, help her to draw this pattern.<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>Triangle()</b> that takes no parameters.Print the pattern as shown in the example.Sample Output:-
*****
****
***
**
*, I have written this Solution Code: static void Triangle(){
System.out.println("*****");
System.out.println("****");
System.out.println("***");
System.out.println("**");
System.out.println("*");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: static void pattern(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j + " ");
}
System.out.println();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: function pattern(n) {
// write code herenum
for(let i = 1;i<=n;i++){
let str = ''
for(let k = 1; k <= i;k++){
if(k === 1) {
str += `${k}`
}else{
str += ` ${k}`
}
}
console.log(str)
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<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 a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: def patternPrinting(n):
for i in range(1,n+1):
for j in range (1,i+1):
print(j,end=' ')
print()
, 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, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int m = 100001;
int main(){
int n,k;
cin>>n>>k;
long long a[n],sum=0;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]<0){
a[i]=-a[i];
}
}
sort(a,a+n);
for(int i=0;i<k;i++){
sum+=a[n-i-1]*a[n-i-1];
}
cout<<sum;
}
, 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, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String[] NK = br.readLine().split(" ");
String[] inputs = br.readLine().split(" ");
int N = Integer.parseInt(NK[0]);
int K = Integer.parseInt(NK[1]);
long[] arr = new long[N];
long answer = 0;
for(int i = 0; i < N; i++){
arr[i] = Math.abs(Long.parseLong(inputs[i]));
}
quicksort(arr, 0, N-1);
for(int i = (N-K); i < N;i++){
answer += (arr[i]*arr[i]);
}
System.out.println(answer);
}
static void quicksort(long[] arr, int start, int end){
if(start < end){
int pivot = partition(arr, start, end);
quicksort(arr, start, pivot-1);
quicksort(arr, pivot+1, end);
}
}
static int partition(long[] arr, int start, int end){
long pivot = arr[end];
int i = start - 1;
for(int j = start; j < end; j++){
if(arr[j] < pivot){
i++;
swap(arr, i, j);
}
}
swap(arr, i+1, end);
return (i+1);
}
static void swap(long[] arr, int i, int j){
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}, 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, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: x,y = map(int,input().split())
arr = list(map(int,input().split()))
s=0
for i in range(x):
if arr[i]<0:
arr[i]=abs(arr[i])
arr=sorted(arr,reverse=True)
for i in range(0,y):
s = s+arr[i]*arr[i]
print(s)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N elements. Find the maximum value of GCD(Arr[i], Arr[j]) where i != j.First line of input contains a single integer N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
2 <= N <= 100000
1 <= Arr[i] <= 100000Print the maximum value of GCD(Arr[i], Arr[j]) where i != j.Sample Input 1
5
2 4 5 2 2
Sample Output 1
2
Explanation: We can select index 1 and index 4, GCD(2, 2) = 2
Sample Input 2
6
4 3 4 1 6 5
Sample Output 2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int result(int a[],int n)
{
int maxe =0;
for(int i=0;i<n;i++)
maxe = Math.max(maxe,a[i]);
int count[]=new int[maxe+1];
for(int i=0;i<n;i++)
{
for(int j=1;j<Math.sqrt(a[i]);j++)
{
if(a[i]%j==0)
{
count[j]++;
if (j != a[i] / j)
count[a[i] / j]++;
}
}
}
for(int i=maxe;i>0;i--)
{
if(count[i]>1)
return i;
}
return 1;
}
public static void main (String[] args) throws IOException{
InputStreamReader i = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(i);
int n = Integer.parseInt(br.readLine());
int a[] = new int[n];
String str = br.readLine();
String[] strs = str.trim().split(" ");
for (int j = 0; j < n; j++)
{
a[j] = Integer.parseInt(strs[j]);
}
System.out.println(result(a,n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of N elements. Find the maximum value of GCD(Arr[i], Arr[j]) where i != j.First line of input contains a single integer N.
Second line of input contains N space separated integers, denoting array Arr.
Constraints:
2 <= N <= 100000
1 <= Arr[i] <= 100000Print the maximum value of GCD(Arr[i], Arr[j]) where i != j.Sample Input 1
5
2 4 5 2 2
Sample Output 1
2
Explanation: We can select index 1 and index 4, GCD(2, 2) = 2
Sample Input 2
6
4 3 4 1 6 5
Sample Output 2
4, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int v[100001]={};
int n;
cin>>n;
for(int i=1;i<=n;++i){
int d;
cin>>d;
for(int j=1;j*j<=d;++j){
if(d%j==0){
v[j]++;
if(j!=d/j)
v[d/j]++;
}
}
}
for(int i=100000;i>=1;--i)
if(v[i]>1){
cout<<i;
return 0;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton wants to take revenge from two apples fallen on his head. So, he applies force F<sub>1</sub> on first apple (mass M<sub>1</sub>) resulting in acceleration of A<sub>1</sub> and F<sub>2</sub> on second apple (mass M<sub>2</sub>) resulting in acceleration of A<sub>2</sub>. Given M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>. Calculate total force applied by him on two apples.
<b>Note:</b> F = M*A is the equation of relation between force, mass and acceleration.First line contains four integers M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub>.
1 <= M<sub>1</sub>, A<sub>1</sub>, M<sub>2</sub>, A<sub>2</sub> <= 100Output total force applied by Newton.INPUT:
1 2 3 4
OUTPUT:
14
Explanation:
Total force is equal to 1*2 + 3*4 = 14., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int m1,a1,m2,a2;
cin >> m1 >> a1 >> m2 >> a2;
cout << (m1*a1)+(m2*a2) << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
st = new StringTokenizer(br.readLine());
String s = st.nextToken();
int curzeroes = 0;
StringBuilder sb = new StringBuilder();
int len = s.length();
for(int i = 0;i<len;i++){
if(s.charAt(i) == '1'){
if(curzeroes == 0){
sb.append("1");
}
else{
curzeroes--;
}
}
else{
curzeroes++;
}
}
for(int i = 0;i<curzeroes;i++){
sb.append("0");
}
if(sb.length() == 0 && curzeroes == 0){
bw.write("-1\n");
}
else{
bw.write(sb.toString()+"\n");
}
bw.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
Sample Output
-1
, I have written this Solution Code: arr = input()
c = 0
res = ""
n =len(arr)
for i in range(n):
if arr[i]=='0':
c+=1
else:
if c==0:
res+='1'
else:
c-=1
for i in range(c):
res+='0'
if len(res)==0:
print(-1)
else:
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 0s and 1s are super cool.
You are given a binary string (string consisting of only zeros and ones). We need to modify the string such that no 0 is followed by a 1. For achieving this, we will find the leftmost occurrence of "01" substring in the string and remove it from the string. We will repeat this operation until there is no substring of the form "01" in the string.
For example, if the initial string is "011010", it will transform in the following manner:
<b>01</b>1010 -> 1<b>01</b>0 -> 10
Find the final remaining string. If the length of remaining string is 0, print -1 instead.The first and the only line of input contains the initial string, S.
Constraints
1 <= |S| <= 300000Output the remaining string. If the length of remaining string is 0, output -1.Sample Input
011010
Sample Output
10
Explanation: Available in the question text.
Sample Input
001101
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;
string s; cin >> s;
stack<int> st;
for(int i = 0; i < (int)s.length(); i++){
if(!st.empty() && s[st.top()] == '0' && s[i] == '1'){
st.pop();
}
else
st.push(i);
}
string res = "";
while(!st.empty()){
res += s[st.top()];
st.pop();
}
reverse(res.begin(), res.end());
if(res == "") res = "-1";
cout << res;
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 pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code: def SortPair(items,n):
items.sort(reverse = True)
return items, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |