Interview Questions

Gundala tarun,cse2020 V. - "#include // Naive method to find a pair in an array with a given sum void findPair(int nums[], int n, int target) { // consider each element except the last for (int i = 0; i < n - 1; i++) { // start from the i'th element until the last element for (int j = i + 1; j < n; j++) { // if the desired sum is found, print it if (nums[i] + nums[j] == target) { printf("Pair found (%d, %d)\n", nums[i], nums[j]); return; } } } // we reach here if the pair is not found printf("Pair not found"); } "See full answer

Saurabh S. - "Basic Approach As BST inorder traversal will result in a sequence of increasing order. Store that order in a vector and get the k-1 index to get the Kth smallest element, similarly access the N-K+1 th element will be the Kth largest element Time Complexity: O(n) Space Complexity O(n) Space Optimized Approach For Kth smallest , start inorder traversal, and keep a counter, decrement the counter when you access the node element. When the counter turns 0 that elementwill be the Kth smal"See full answer

Vipin G. - "// Function to delete a node from BST. public static Node deleteNode(Node root, int X) { // code here. if(root == null) return root; if(X root.data){ root.right = deleteNode(root.right, X); return root; } if(root.left == null){ return root.right; }else if(root.right == null){ "See full answer

Ugo C. - "function constructTree(n, matrix) { let parent = []; let child = []; let root = null; for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { if (matrixi === 1) { parent.push(i); child.push(j); } } } for (let i = 0; i < n; i++) { if (parent.indexOf(i) === -1) { root = i; } } let node = new Node(root); for (let i = 0; i < n; i++) { if (i !== root) { constructTreeUtil(node, parent[i], child[i]); } } return node; }"See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Nishant R. - "Recursively get the left height and right height of the sub-tree, and then check if it is balanced at this node, return false."See full answer

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer

Prashant Y. - "The height of a binary tree is the maximum number of edges from the root node to any leaf node. To calculate the height of a binary tree, we can use a recursive approach. The basic idea is to compare the heights of the left and right subtrees of the root node, and return the maximum of them plus one."See full answer

Tushar A. - "class Node { int val; Node left, right; Node(int v) { val = v; left = right = null; } } class BinaryTree { Node root1, root2; boolean identicalTrees(Node a, Node b) { if (a == null && b == null) return true; if (a != null && b != null) return (a.val == b.val && identicalTrees(a.left, b.left) && identicalTrees(a.right, b.right)); "See full answer

Denis F. - "You may backtracking it but I don't see what relation this matrix represents?"See full answer

Brajesh J. - "dp"See full answer

Rabiul H. - "htrhtrd"See full answer