Data Structures & Algorithms Interview Questions

Mourya C. - "I try to solve this initially using quick select where will take a pivot element and position the remaining elements and check if the current index is answer or not and continue the same but it requires o(n*n), but interviewee is expecting the best from me, so at the end i tried solving using heaps where will check the difference between k and n-k to use min or max heap after that we will heap the array, and will keep popping the element k-1 and return the peek one which leads to answer."See full answer

Shar N. - "standard answer for this."See full answer

Susmita S. - "Function that transforms each elements in a collection and returns new collection with transformed elements"See full answer

Shivam P. - " First, sort the array in ascending order. This ensures that we can easily check the triangle inequality condition. Use a loop to iterate through the array. For each triplet of consecutive elements, check if they satisfy the triangle inequality condition a+b>ca+b>c. As soon as you find a valid tuple, return it. If no valid tuple is found, return null. This approach is efficient with a time complexity of O(nlog⁡n)O(nlogn) due to the sorting step, followed by a linear scan of the array"See full answer

Iris F. - "#include #include #include using namespace std; void printComs(int prev, int start, int end, int target) { if (start >= end) return; while (start target) { end--; } else { st"See full answer

Shaxboz A. - "#include #include using namespace std; int main() { int n; cin >> n; int a[n]; for(int i=0; i=0 and a[i]<=2) { sort(a[0], a[n]); } } cout << "After sorting array: "; for(int i=0; i<n; i++) { cout << a[i] << " "; } }"See full answer

Yash S. - "Run length encoding. This will preserve order of values in vector."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

Aniket G. - "public class BoggleBoard { public static List findWords(char board, Set dictionary) { int rows = board.length; int cols = board[0].length; boolean visited = new booleanrows; int directions = {{1,0}, {-1,0}, {0,1}, {0,-1}}; List result = new ArrayList(); for(int i=0; i<rows; i++) { for(int j=0; j<cols; j++) { dfs(board, visited, i, j, dictionary, "", result, dire"See full answer