Interview Questions

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

Flora K. - "USING RECURSION"See full answer

Rabiul H. - "htrhtrd"See full answer

Kamlesh S. - "min-num = arr[0] max-num = arr[0] for (i=1; i max-num) { max-num = arr[i]; } } System.out.println("Minimum = " + min-num); System.out.println("Maximum = " + max-num);"See full answer

Anmol R. - "The example given is wrong. The 2nd test case should have answer 3, as we can get to 6 by using 3 coins of denomination 2."See full answer

Shubhendu K. - "// array is sorted in non-increasing order // low = 0, high = arr.length - 1 int countOnes(int[] arr, int low, int high) { if (high >= low) { int mid = low + (high - low) / 2; if ((mid == high || arr[mid + 1] == 0) && (arr[mid] == 1)) { return mid + 1; } if (arr[mid] == 1) { return countOnes(arr, (mid + 1), high); } return countOnes(arr, low, (mid - 1)); } return 0; }"See full answer

Ankur T. - "package mycoding.tryexponent.com; import java.util.Scanner; public class FirstAndLastOccurenceOfaNumber { public static void main(String[] arg) { System.out.println("Enter 'a' to run the programme:"); System.out.println("Enter 't' to terminate the programme:"); Scanner in=new Scanner(System.in); String inputtedString=in.nextLine(); while(!"t".equalsIgnoreCase(inputtedString)) { System.out.println("enter a number to find its 1st nd last occurence"); int number=in.nextInt(); int[] array=new i"See full answer

Ugo C. - "function countRotations(arr, low, high) { if (high low && arr[mid] arr[mid]) { return countRotations(arr, low, mid - 1); } return countRotations(arr, mid + 1, high); } "See full answer

Satyam S. - "Binary search is commonly used for searching elements in a sorted array. Most searching algorithms take O(n) time, but binary search operates in O(log(n)) time complexity. function binarySearch(arr, target) { let first = 0; let last = arr.length - 1; // Adjusted to correctly represent the last index while (first target) { last = mid - 1; } "See full answer

Ashesh S. - "Steps: Validate K to be less than the number of elements in an array. Sort the array in ascending order. Get the K'th smallest element by array[k-1]."See full answer

Anonymous Narwhal - "As the array is sorted, would it best to use binary search to find the one the triplet and just navigate from it."See full answer

Hinata T. - "bool isConsecutive(int arr[], int n) { // base case if (n max) { max = arr[i]; } } // for an array to contain consecutive integers, the difference between // the maximum and minimum element in it should be exactly \n-1\ if (max - min != n - 1) { return false; } // create an empty set (we can also use a visit"See full answer