pLft = new NODE(1);\n\n root->pRgt = new NODE(4);\n\n char str[] = \"\";\n\n Serialize(str, root);\n\n cout<
val<pLft->val<pRgt->val<\n\n#include\n\n#include\n\nusing namespace std;\n\n\n\nbool TwoSum(int arr[], int len, int skipindex) {\n\n\n","file_path":"problem/Array/Find the max item A[i], such that A[i]=A[x]+A[y].cpp","rank":88,"score":17.058967243610713},{"content":"*/\n\n#include\n\nusing namespace std;\n\n\n\n\n\n\n\nbool IsDuplicate(const string str) {\n\n\n\n int strlen = str.size();\n\n\n\n if(strlen < 4) return false;\n\n\n\n int len = 1; //length of the duplicate substring\n\n int start = 0;\n\n for (int i = 1; i< strlen; ++i) {\n\n\n\n if (str[start] == str[i]) {\n\n\n\n start++;\n\n if(start >= len) start = 0; //reach an end of a duplicate substring, back to beginning\n","file_path":"problem/String/Kduplicates.cpp","rank":89,"score":16.917550227629544},{"content":"\n\nuse two sets, one is to record all visited index in the array, another one is to record the visited index in current possible cycle\n\nO(n) time, O(n) space\n\n\n\n*/\n\n\n\n\n\n#include\n\n#include\n\nusing namespace std;\n\n\n\nbool DetectCycle1(int arr[], int len) {\n\n \n\n int slow = 0;\n\n int fast = 0;\n\n int cur = 0;\n\n \n\n while (cur < len) {\n\n \n\n if (arr[slow] >= 0 && arr[slow] < len) {\n","file_path":"problem/Array/Detect cycle in an array.cpp","rank":90,"score":16.758783001097143},{"content":"\n\n if(local) {\n\n return local;\n\n\t\t} else{\n\n node = parent;\n\n parent = node->parent;\n\n level--;\n\n }\n\n }\n\n return NULL; \n\n}\n\n\n\nint main(){\n\n\n\n Node *root = new Node(0);\n\n root->left = new Node(1);\n\n root->left->parent = root;\n\n root->right = new Node(2);\n\n root->right->parent = root;\n\n root->left->left = new Node(3);\n","file_path":"problem/Tree/Immediate right neighbor.cpp","rank":91,"score":16.717273349583355},{"content":"/*\n\n\n\nsolutuion: calculate horizontal distances from root for all nodes. If two nodes have the same horizontal distance, then they are on same vertical line. \n\nHD for root is 0, a right edge (edge connecting to right subtree) is considered as +1 horizontal distance and a left edge is considered as -1 horizontal distance. \n\n\n\nDo an inorder traversal and calcuate the horizontal distance recursively.\n\n\n\nO(n) time, O(n) space\n\n\n\n*/\n\n\n\n#include\n\n#include\n\nusing namespace std;\n\n\n","file_path":"problem/Tree/Vertical sum in a binary tree.cpp","rank":92,"score":16.683155915992955},{"content":"/*\n\nSieve of Eratosthenes algorithm\n\n*/\n\n\n\n/*\n\n\n\nsolution: start from smallest primer number p, print p and mark 2p, 3p, 4p, ...\n\nprint the first number q greater than p without mark if it exists, then repeat the steps above for q.\n\n\n\n*/\n\n\n\n\n\n#include \n\nusing namespace std;\n\n\n\n\n\nvoid FlagMultiples(bool arr[], int a, unsigned int num) {\n\n int i = 2;\n\n int temp = i * a;\n\n while ( temp <= num ) {\n","file_path":"problem/SieveofEratosthenes.cpp","rank":93,"score":16.682443619178972},{"content":"*/\n\n\n\n/*\n\n\n\nsolution: define the vowel checking rule correctly. Then just one scan.\n\nO(n) time, O(1) space\n\n\n\n*/\n\n#include\n\n#include\n\n#include\n\nusing namespace std;\n\n\n\nbool checkBasicVowel(char c) {\n\n\n\n if (c == 'a' || c =='e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' ||c == 'U') {\n\n return true;\n\n\n\n } else {\n\n\n","file_path":"problem/String/Vowel product.cpp","rank":94,"score":16.664347476311875},{"content":"\n\n#include\n\n#include\n\nusing namespace std;\n\n\n\nbool TransformString(string str1, string str2) {\n\n\n\n int len1 = str1.size();\n\n int len2 = str2.size();\n\n\n\n if (abs(len1 - len2) > 1) return false;\n\n\n\n if (len1 > len2) return TransformString(str2, str1);\n\n\n\n\t//replace\t\n\n if (len1 == len2) {\n\n\n\n int countdiff = 0;\n\n\n\n for (int i = 0 ; i < len1; ++i) {\n","file_path":"problem/String/Edit Distance one.cpp","rank":95,"score":16.567559041272233},{"content":"/*\n\nGiven an array of sorted integers where each one is in the range from 0 to m-1 and m > n,\n\nwhere n is the length of this array.\n\nFind the smallest number that is missing from the array.\n\n*/\n\n\n\n/*\n\nsolution: modified binary search. \n\nO(logn) time, O(1) space\n\n*/\n\n\n\n#include\n\nusing namespace std;\n\n\n\nint FindFirstMissing(int arr[], int begin, int end) {\n\n \n\n //end+1 is the number larger than the last element of array\n\n if (begin > end)\n\n return end + 1;\n\n \n","file_path":"problem/SortingSearch/Smallestmissingnumber.cpp","rank":96,"score":16.234300952581133},{"content":" stk.pop();\n\n }\n\n }\n\n}\n\n \n\n//preorder\n\nvoid PrevOrderTravel(NODE* root) {\n\n assert(root);\n\n \n\n stack stk;\n\n stk.push(root);\n\n \n\n while (!stk.empty()) {\n\n \n\n NODE* top = stk.top();\n\n stk.pop();\n\n \n\n cout<val<pRgt != NULL) stk.push(top->pRgt);\n","file_path":"problem/Tree/PreInPostorder travel of bT with stack.cpp","rank":97,"score":15.901693231718374},{"content":"/*\n\nFind the last k digits of m^n, where m, n is positive integer, n may be very large.\n\n*/\n\n\n\n/*\n\n\n\nsolution: big number + recursion\n\nO((log n) * k * k) time, O(k*k) space\n\n*/\n\n\n\n#include\n\n#include\n\n#include\n\nusing namespace std;\n\n\n\nstring MultiplyArrays(string num1, string num2, int k) {\n\n int N = num1.length(), M = num2.length();\n\n string res(N+M, '0');\n\n for (int i = N - 1; i >= 0; --i) {\n\n int carry = 0;\n","file_path":"problem/Math/Calculate (m^n)%(10^k).cpp","rank":98,"score":15.607745224466276},{"content":"","file_path":"problem/Quadtree.cpp","rank":99,"score":15.453547276372685}],"string":"[\n {\n \"content\": \"struct Node {\\n\\n int val;\\n\\n Node* plft;\\n\\n Node* prgt;\\n\\n\\n\\n\\tNode(int v) : val(v), plft(NULL), prgt(NULL) {}\\n\\n};\\n\\n\\n\\n\\n\\n//need finish this version\\n\\n\\n\\nint main() {\\n\\n\\n\\n\\t/*\\n\\n\\tNode *root = new Node(1);\\n\\n\\troot->plft = new Node(2);\\n\\n\\troot->prgt = new Node(3);\\n\\n\\troot->plft->plft = new Node(4);\\n\\n\\troot->plft->prgt = new Node(5);\\n\\n\\troot->prgt->plft = new Node(6);\\n\",\n \"file_path\": \"problem/Tree/Printing all nodes.cpp\",\n \"rank\": 0,\n \"score\": 115952.27447917088\n },\n {\n \"content\": \"struct Node {\\n\\n int val;\\n\\n Node *plft;\\n\\n Node *prgt;\\n\\n \\n\\n Node(int v) : val(v), plft(NULL), prgt(NULL) {}\\n\\n};\\n\\n\\n\\n\\n\\n\\n\\n\\n\\nNode *SecondLargestNode(NODE *root) {\\n\\n\\n\\n\\n\\n if (root == NULL || root->plft == NULL && root->prgt == NULL) return NULL;\\n\\n\\n\\n NODE *iter = root;\\n\\n NODE *parent = NULL;\\n\\n \\n\\n while (iter->prgt != NULL) {\\n\",\n \"file_path\": \"problem/Tree/Second largest node in BST.cpp\",\n \"rank\": 1,\n \"score\": 112531.14414397393\n },\n {\n \"content\": \"struct Node {\\n\\n\\n\\n int val\\n\\n Node *plft;\\n\\n TreeNode *prgt;\\n\\n \\n\\n Node(int v) : val(v), plft(NULL), prgt(NULL) {}\\n\\n\\n\\n};\\n\\n\\n\\nint Height(Node* root) {\\n\\n\\n\\n if (root == NULL) {\\n\\n return 0;\\n\\n }\\n\\n\\n\\n int leftheight = Height(root->plft);\\n\\n int rightheight = Height(root->prgt);\\n\\n\\n\\n if (leftheight > rightheight) {\\n\",\n \"file_path\": \"problem/Tree/Longest distance between two nodes.cpp\",\n \"rank\": 2,\n \"score\": 112531.14414397393\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* pLft;\\n\\n NODE* pRgt;\\n\\n \\n\\n NODE(int n) : val(n), pLft(NULL), pRgt(NULL) {}\\n\\n};\\n\\n\\n\\nvoid GetRightMostBTInner(NODE* Node, vector &vec, int Lev) {\\n\\n\\n\\n if (Node == NULL) return;\\n\\n \\n\\n if (Lev >= vec.size()) {\\n\\n \\n\\n vec.push_back(Node);\\n\\n cout<val;\\n\\n }\\n\\n\\t\\n\\n \\n\\n GetRightMostBTInnerNode->pRgt, vec, Lev+1);\\n\",\n \"file_path\": \"problem/Tree/Find the right most node in each level of a binary tree.cpp\",\n \"rank\": 4,\n \"score\": 109383.88740781978\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n vector vec;\\n\\n \\n\\n NODE(int n) : val(n), vec() {}\\n\\n};\\n\\n \\n\\nvoid SerialInner(NODE* node, char*& p) {\\n\\n\\n\\n\\t\\n\\n itoa(node->val, p, 10);\\n\\n p += strlen(p);\\t\\n\\n int num = node->vec.size();\\n\\n itoa(num, p, 10);\\n\\n p += strlen(p);\\n\\n\\t\\n\\n if (node->vec.size() == 0) return;\\n\\n \\n\\n for (vector::iterator it = node->vec.begin(); it != pNode->vec.end(); it++) {\\n\\n \\n\",\n \"file_path\": \"problem/Tree/SerializeDeSerialize an arbitrary nodes tree.cpp\",\n \"rank\": 5,\n \"score\": 109383.88740781978\n },\n {\n \"content\": \"struct node {\\n\\n double _x, _y;\\n\\n node(double x, double y): _x(x), _y(y){}\\n\\n node(): _x(0.0), _y(0.0){}\\n\\n};\\n\\n\\n\\ndouble Distance(node x, node target) {\\n\\n\\treturn pow(x._x - target._x, 2) + pow(x._y - target._y,2);\\n\\n}\\n\\n\\n\\nnode target(0,0);\\n\\n\\n\",\n \"file_path\": \"problem/Heap/TwoDtopk.cpp\",\n \"rank\": 6,\n \"score\": 107116.52377831328\n },\n {\n \"content\": \"struct Node {\\n\\n int val;\\n\\n Node *left;\\n\\n Node *right;\\n\\n Node *parent;\\n\\n \\n\\n Node(int v) : val(v), left(NULL), right(NULL), parent(NULL) {}\\n\\n};\\n\\n\\n\\n\\n\\nNode *FindNeighbor( Node *node, int level ) {\\n\\n\\n\\n if( node == NULL ) return NULL;\\n\\n if( level == 0 ) return node;\\n\\n\\n\\n Node *left = FindNeighbor(node->left,level+1);\\n\\n //always return left child at first such that this node is the immediate right neighbor\\n\\n if( left ) { \\n\\n\\t\\treturn left;\\n\\n\\t} else { \\n\",\n \"file_path\": \"problem/Tree/Immediate right neighbor.cpp\",\n \"rank\": 7,\n \"score\": 105134.336527965\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* plft;\\n\\n NODE* prgt;\\n\\n \\n\\n NODE(int n) : val(n), plft(NULL), prgt(NULL) {}\\n\\n};\\n\\n\\n\\nvoid ShiftUp(NODE *node) {\\n\\n\\n\\n if(!node) return; \\n\\n \\n\\n NODE *largest = node;\\n\\n\\n\\n\\t //find the largest node\\n\\n if (node->plft != NULL && largest->val < node->plft->val) { \\n\\n\\n\\n largest = node->plft; \\n\\n\\n\\n }\\n\",\n \"file_path\": \"problem/Tree/Heapify a binary tree.cpp\",\n \"rank\": 8,\n \"score\": 105134.336527965\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* pLft;\\n\\n NODE* pRgt;\\n\\n \\n\\n NODE(int n) : val(n), pLft(NULL), pRgt(NULL) {}\\n\\n};\\n\\n\\n\\n/*\\n\\nHere is the inorder iterator.\\n\\n*/\\n\\n\\n\",\n \"file_path\": \"problem/Tree/Iterator for binary tree.cpp\",\n \"rank\": 9,\n \"score\": 105134.336527965\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* pLft;\\n\\n NODE* pRgt;\\n\\n int size; //the size of the subtree rooted at that node\\n\\n NODE* pParent;\\n\\n \\n\\n NODE(int n, int sz = 1) : val(n), size(sz), pLft(NULL), pRgt(NULL), pParent(NULL) {}\\n\\n \\n\\n void setLeft(NODE* node) {\\n\\n if (node==NULL) return;\\n\\n pLft = node;\\n\\n node->pParent = this;\\n\\n }\\n\\n \\n\\n void setRight(NODE* node) {\\n\\n if (node==NULL) return;\\n\\n pRgt = node;\\n\\n node->pParent = this;\\n\\n }\\n\",\n \"file_path\": \"problem/Tree/Statistical binary search tree.cpp\",\n \"rank\": 10,\n \"score\": 103250.96983060292\n },\n {\n \"content\": \"struct Node {\\n\\n int val;\\n\\n Node *plft;\\n\\n Node *prgt;\\n\\n \\n\\n Node(int v) : val(v), plft(NULL), prgt(NULL) {}\\n\\n};\\n\\n\\n\\n\\n\\nvoid GetHorizontalDistance(Node *root, int hd, unordered_map &hdmp) { \\n\\n\\n\\n \\n\\n if (root == NULL) return;\\n\\n \\n\\n \\n\\n GetHorizontalDistance(root->plft, hd - 1, hdmp);\\n\\n \\n\\n \\n\\n int prevsum = (hdmp.find(hd) == hdmp.end()) ? 0 : hdmp.find(hd)->second;\\n\\n\\n\",\n \"file_path\": \"problem/Tree/Vertical sum in a binary tree.cpp\",\n \"rank\": 11,\n \"score\": 103250.96983060292\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* pLft;\\n\\n NODE* pRgt;\\n\\n \\n\\n NODE(int n) : val(n), pLft(NULL), pRgt(NULL) {}\\n\\n};\\n\\n\\n\\n//postorder\\n\\nvoid PostOrderTravel(NODE* root) {\\n\\n assert(root);\\n\\n \\n\\n stack stk;\\n\\n NODE* cur = root;\\n\\n while (cur != NULL) {\\n\\n \\n\\n stk.push(cur);\\n\\n cur = cur->pLft;\\n\\n }\\n\\n \\n\",\n \"file_path\": \"problem/Tree/PreInPostorder travel of bT with stack.cpp\",\n \"rank\": 12,\n \"score\": 101459.21347241984\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* pLft;\\n\\n NODE* pRgt;\\n\\n \\n\\n NODE(int n) : val(n), pLft(NULL), pRgt(NULL) {}\\n\\n};\\n\\n \\n\\nvoid SerializeInner(char * &p, NODE* node) {\\n\\n if (node == NULL) {\\n\\n *p++ = '#';\\n\\n *p++ = ' ';\\n\\n return;\\n\\n }\\n\\n //char * itoa ( int value, char * str, int base );\\n\\n //Converts an integer value to a null-terminated string using \\n\\n //the specified base and stores the result in the array given\\n\\n //by str parameter.\\n\\n\\t\\n\\n itoa(node->val, p, 10);\\n\",\n \"file_path\": \"problem/Tree/SerializeDeSerialize a binary tree.cpp\",\n \"rank\": 13,\n \"score\": 101459.21347241984\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* next;\\n\\n \\n\\n NODE(int x) : val(x), next(NULL) {}\\n\\n};\\n\\n\\n\\n//insert x after p\\n\\nvoid Insert(NODE* p, int x) {\\n\\n NODE* tmp = p->next;\\n\\n NODE* newnode = new NODE(x);\\n\\n p->next = newnode;\\n\\n newnode->next = tmp;\\n\\n}\\n\\n \\n\\nvoid InsertToCycleLink(NODE* node, int x) {\\n\\n assert(node != NULL);\\n\\n \\n\\n NODE* iter = node;\\n\\n NODE* big = iter; \\n\",\n \"file_path\": \"problem/LinkedList/Insert into a Cyclic Sorted List.cpp\",\n \"rank\": 14,\n \"score\": 101459.21347241984\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* Lft;\\n\\n NODE* Rgt;\\n\\n \\n\\n NODE(int n) : val(n), Lft(NULL), Rgt(NULL) {}\\n\\n};\\n\\n \\n\\nbool BinTreeSwitchEqualInner(NODE* node1, NODE* node2, int& switch) {\\n\\n switch = 0;\\n\\n\\t\\n\\n if (node1 == NULL || node2 == NULL) return node1 == NULL && node2 == NULL;\\n\\n \\n\\n if (node1->val != node2->val) return false;\\n\\n \\n\\n int leftswitch, rightswitch;\\n\\n \\n\\n if (BinTreeSwitchEqualInner(node1->Lft, node2->Lft, leftswitch) && \\n\\n BinTreeSwitchEqualInner(node1->Rgt, node2->Rgt, rightswitch)) {\\n\\n \\n\",\n \"file_path\": \"problem/Tree/Switch binary tree to make equal.cpp\",\n \"rank\": 15,\n \"score\": 101459.21347241984\n },\n {\n \"content\": \"struct NODE {\\n\\n int val;\\n\\n NODE* pLft;\\n\\n NODE* pRgt;\\n\\n \\n\\n NODE(int n) : val(n), pLft(NULL), pRgt(NULL) {}\\n\\n};\\n\\n\\n\\n \\n\\nint GetLargestBSTInner(NODE* Node, int& minbound, int& maxbound, NODE*& maxbstroot, int& maxbstnum) {\\n\\n \\n\\n assert(Node);\\n\\n \\n\\n minbound = Node->val;\\n\\n maxbound = Node->val;\\n\\n \\n\\n int result = 1; //count root\\n\\n bool bivalid = false;\\n\\n if (Node->pLft != NULL) {\\n\\n \\n\",\n \"file_path\": \"problem/Tree/Find the largest BST sub tree in BT.cpp\",\n \"rank\": 16,\n \"score\": 99752.5420206213\n },\n {\n \"content\": \"struct node {\\n\\n int a,b,val; //element index and their value sum\\n\\n node(int _a,int _b,int v): a(_a),b(_b),val(v) {}\\n\\n\\t\\n\\n};\\n\\n\\n\",\n \"file_path\": \"problem/Heap/Kth maximal element of sum of two arrays.cpp\",\n \"rank\": 17,\n \"score\": 99752.5420206213\n },\n {\n \"content\": \"struct TNode {\\n\\n int val;\\n\\n TNode* plft;\\n\\n TNode* prgt;\\n\\n TNode* parent;\\n\\n\\n\\n TNode(int v) : val(v), plft(NULL), prgt(NULL), parent(NULL) {}\\n\\n};\\n\\n\\n\\n//print node below given node \\n\\nvoid PrintDown(TNode * node, int k) {\\n\\n\\n\\n if (node == NULL || k < 0) return;\\n\\n\\n\\n if (k == 0) {\\n\\n cout<val<plft, k -1);\\n\\n PrintDown(node->prgt, k -1);\\t\\n\\n }\\n\",\n \"file_path\": \"problem/Tree/Printing all nodes.cpp\",\n \"rank\": 18,\n \"score\": 97779.65219680007\n },\n {\n \"content\": \"\",\n \"file_path\": \"problem/Quadtree.cpp\",\n \"rank\": 19,\n \"score\": 92779.9945778869\n },\n {\n \"content\": \"struct TreeNode {\\n\\n int val;\\n\\n vector children; \\n\\n TreeNode(int n) : val(n), children() {}\\n\\n};\\n\\n\\n\\nint MaxPathSumInner(TreeNode *node, int &maxSoFar) {\\n\\n if (node == NULL) return 0;\\n\\n\\tint maxPath = node->val;\\n\\n\\tint tempPath = INT_MIN;\\n\\n\\tvector::iterator it = node->children.begin();\\n\\n\\tfor (; it != node->children.end(); it++) {\\n\\n\\t\\tint maxchild = MaxPathSumInner(*it, maxSoFar);\\n\\n\\t\\t\\n\\n\\t\\tif (maxchild > 0) maxPath += maxchild;\\n\\n\\t\\tif (maxchild > tempPath) tempPath = maxchild;\\n\\n\\t}\\n\\n\\tif (maxPath > maxSoFar) maxSoFar = maxPath;\\n\\n\\tint res = node->val;\\n\\n\\treturn max(res, res + tempPath); //note path definition\\n\",\n \"file_path\": \"problem/Tree/Maximalpathsum.cpp\",\n \"rank\": 20,\n \"score\": 90690.99953321335\n },\n {\n \"content\": \"struct Racer {\\n\\n\\n\\n\\n\\n Racer(int ide, int starte, int ende) {\\n\\n id = ide;\\n\\n start = starte;\\n\\n end = ende;\\n\\n startrank = 0;\\n\\n score = 0;\\n\\n }\\n\\n Racer() {\\n\\n id = -1;\\n\\n start = -1;\\n\\n end = -1;\\n\\n startrank = 0;\\n\\n score = 0;\\n\\n }\\n\\n\\n\\n int id, start, end, score, startrank;\\n\\n};\\n\",\n \"file_path\": \"problem/Racescore.cpp\",\n \"rank\": 21,\n \"score\": 54179.822181620446\n },\n {\n \"content\": \"struct Event {\\n\\n Event(int _h, int _x, int _type) \\n\\n : height(_h), x(_x), type(_type) {}\\n\\n enum Type{START, END};\\n\\n int height;\\n\\n int x;\\n\\n int type;\\n\\n};\\n\\n/*\\n\\nsort the event from begin to end by the following order:\\n\\nthe event with smaller x axis value, with smaller type,\\n\\nif both event are START, with larger height, or if both\\n\\nevent are END, with smaller height. \\n\\n\\n\\n*/\\n\\nbool CmpEvent(const Event* lhs, const Event* rhs) {\\n\\n if (lhs->x == rhs->x) {\\n\\n if (lhs->type == rhs->type) {\\n\\n if (lhs->type == Event::START) {\\n\\n return lhs->height > rhs->height;\\n\",\n \"file_path\": \"problem/Skyline.cpp\",\n \"rank\": 22,\n \"score\": 54179.822181620446\n },\n {\n \"content\": \"\",\n \"file_path\": \"problem/Onlineuser.cpp\",\n \"rank\": 23,\n \"score\": 54179.822181620446\n },\n {\n \"content\": \"\",\n \"file_path\": \"problem/Groupcontact.cpp\",\n \"rank\": 24,\n \"score\": 54179.822181620446\n },\n {\n \"content\": \"\",\n \"file_path\": \"problem/Onlineuser.cpp\",\n \"rank\": 25,\n \"score\": 54179.822181620446\n },\n {\n \"content\": \"struct Compareto {\\n\\n bool operator() (const Cell *a, const Cell *b) {\\n\\n return a->height > b->height;\\n\\n }\\n\\n};\\n\\n\\n\\nconst int m = 3;\\n\\nconst int n = 3;\\n\\npriority_queue, Compareto> minheap; \\n\\nvector > visited(m, vector(n, false));\\n\\nvector > mat(m, vector(n, 0));\\n\\n\\n\\n\\n\\n\\n\\nvoid Check(int x, int y, int lowest, int &result) {\\n\\n\\n\\n if(x < 0 || x >= m || y < 0 || y >= n || visited[x][y]) return;\\n\\n if(mat[x][y] < lowest) result += lowest - mat[x][y];\\n\\n visited[x][y] = true;\\n\\n minheap.push(new Cell(x,y, max(lowest, mat[x][y])));\\n\",\n \"file_path\": \"problem/Heap/Waterin2dhistogram.cpp\",\n \"rank\": 26,\n \"score\": 53074.507745440525\n },\n {\n \"content\": \"struct Cell {\\n\\n\\n\\n int x;\\n\\n int y;\\n\\n int height;\\n\\n\\tCell(): x(0), y(0), height(0) {}\\n\\n Cell(int _x,int _y,int _height): x(_x), y(_y), height(_height) {}\\n\\n\\n\\n};\\n\\n\\n\",\n \"file_path\": \"problem/Heap/Waterin2dhistogram.cpp\",\n \"rank\": 27,\n \"score\": 53074.507745440525\n },\n {\n \"content\": \"struct GreaterNote {\\n\\n\\n\\n bool operator() (const node a, const node b) {\\n\\n return Distance(a, target) < Distance(b, target);\\n\\n }\\n\\n};\\n\\n\\n\\n\\n\\n\\n\\nvector GetKClosestNode(vector nodes, int k) {\\n\\n priority_queue, GreaterNote > q; //build the max heap\\n\\n size_t size = nodes.size();\\n\\n if (size <= k) return nodes;\\n\\n for (size_t i = 0; i < k; i++ ){\\n\\n q.push(nodes[i]);\\n\\n }\\n\\n size_t num = k;\\n\\n\\t\\n\\n while (num < size) {\\n\\n node top = q.top();\\n\",\n \"file_path\": \"problem/Heap/TwoDtopk.cpp\",\n \"rank\": 28,\n \"score\": 51033.41905068699\n },\n {\n \"content\": \"/////////////////////////////////////////// \\n\\nstruct TOKEN {\\n\\n bool bnull;\\n\\n int val;\\n\\n TOKEN(bool b, int n) : bnull(b), val(n) {}\\n\\n};\\n\\n \\n\\nTOKEN GetToken(const char* &p) {\\n\\n while(*p == ' ') p++;\\n\\n \\n\\n if (*p < '0' || *p > '9') {\\n\\n \\n\\n p++;\\n\\n return TOKEN(true, 0);\\n\\n }\\n\\n \\n\\n int val = 0;\\n\\n while (*p >= '0' && *p <= '9') {\\n\\n \\n\\n val = 10*val + *p++ - '0';\\n\\n \\n\",\n \"file_path\": \"problem/Tree/SerializeDeSerialize a binary tree.cpp\",\n \"rank\": 29,\n \"score\": 49190.89479681765\n },\n {\n \"content\": \"struct GreaterNote {\\n\\n\\n\\n bool operator() (const node a, const node b) {\\n\\n return a.val < b.val;\\n\\n }\\n\\n};\\n\\n\\n\\nint FindKthSum1(int A[],int m,int B[],int n,int k) {\\n\\n\\n\\n\\tpriority_queue, GreaterNote > q; //build the max heap\\n\\n\\tq.push(node(0,0,A[0]+B[0]));\\n\\n\\tset > visited;\\n\\n\\tvisited.insert(pair(0,0));\\n\\n\\twhile(!q.empty()) {\\n\\n\\t\\n\\n\\t\\tnode t = q.top();\\n\\n\\t\\tq.pop();\\n\\n\\t\\tk--;\\n\\n\\t\\tif(k == 0) return t.val;\\n\\n\\t\\t\\n\",\n \"file_path\": \"problem/Heap/Kth maximal element of sum of two arrays.cpp\",\n \"rank\": 30,\n \"score\": 47519.303438591596\n },\n {\n \"content\": \"using namespace std;\\n\\n\\n\\nbool Prob() {\\n\\n\\n\\n int i = rand();\\n\\n if (i % 2 == 0) {\\n\\n\\n\\n return true;\\n\\n\\n\\n } else {\\n\\n\\n\\n return false;\\n\\n\\n\\n }\\n\\n}\\n\\n\\n\\nbool ProbAny(double probability, bool value ) {\\n\\n\\n\\n if (probability < 0.5) {\\n\\n\\t\\n\",\n \"file_path\": \"problem/Tree/Output true.cpp\",\n \"rank\": 31,\n \"score\": 40305.559503098884\n },\n {\n \"content\": \"/*\\n\\n\\n\\nCalculate the last digit of a^b, where a,b are positive. Use map to store the last digit.\\n\\nO(logb) time, O(10) space\\n\\n\\n\\n*/\\n\\n\\n\\n#include\\n\\n#include