Data Structures & Algorithms Interview Questions

Khushbu R. - "Let’s say the matrix is m x n (i.e., m rows and n columns). Start from the top-right corner of the matrix. Move left if you see a 1. Move down if you see a 0. Keep track of the row index where you last saw the leftmost 1 — that row has the most 1s. public class MaxOnesRow { public static int rowWithMostOnes(int matrix) { int rows = matrix.length; int cols = matrix[0].length; int maxRowIndex = -1; int j = cols - 1; /"See full answer

Srihitha J. - "my answer: void* memcpy(void* dest, const void* src, size_t n) { unsigned char* uDest = static_cast(dest); const unsigned char* ucSrc = static_cast(src); for(size_t i= 0; i(dest); const unsigned c"See full answer

Akash C. - " from typing import Optional class Node: def init(self, val: int, prev: Optional['Node'] = None, next: Optional['Node'] = None): self.val = val self.prev = prev self.next = next def split(head): if not head or not head.next: return head slow = head fast = head.next while fast and fast.next: slow = slow.next fast = fast.next.next mid = slow.next slow.next = None if mid: mid.prev = None "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

Ramachandra N. - "def mergeTwoListsRecursive(l1, l2): if not l1 or not l2: return l1 or l2 if l1.val < l2.val: l1.next = mergeTwoListsRecursive(l1.next, l2) return l1 else: l2.next = mergeTwoListsRecursive(l1, l2.next) return l2 "See full answer

Vaibhav C. - "To determine if a graph is not a tree, you can check for the following conditions: Presence of cycles: A graph is not a tree if it contains cycles. In a tree, there should be exactly one unique path between any two vertices. If you can find a cycle in the graph, it cannot be a tree. Insufficient number of edges: A tree with N vertices will have exactly N-1 edges. If the graph has fewer or more than N-1 edges, then it is not a tree. Disconnected components: A tree is a connected graph, m"See full answer

Reno S. - "import Foundation func spiralCopy(inputMatrix: [[Int]]) -> [Int] { let arr = inputMatrix var top = 0, down = arr.count - 1 var left = 0, right = arr[0].count - 1 if top == down && left == right { return arr[top] } var ans: [Int] = [] while top <= down && left <= right { for i in left..<right { ans.append(arrtop) } for i in top..<down { ans.append(arri) } fo"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

Michael B. - "Problem: Given a modified binary tree, where each node also has a pointer to it's parent, find the first common ancestor of two nodes. Answer: As it happens, the structure that we're looking at is actually a linked list (one pointer up), so the problem is identical to trying to find if two linked lists share a common node. How this works is by stacking the two chains of nodes together so they're the same length. chain1 = node1 chain2= node2 while True: chain1 = chain1.next chain2=chain"See full answer

B. T. - "You are given a string S and a number K as input, and your task is to print S to console output considering that, at most, you can print K characters per line. Example: S = "abracadabra sample" K = 11 Output: abracadabra sample Note that this problem requires the interviewee gather extra requirements from the interviewer (e.g. do we care about multiple white spaces? what if the length of a word is greater than K, ...)"See full answer

Jack99 - "Node* getLeftMostChild(Node* node){ while(node->left){ node = node->left; } return node; } Node* findInOrderSuccessor( Node *inputNode ) { int val = inputNode->key; if(inputNode->right){ return getLeftMostChild(inputNode->right); }else{ inputNode = inputNode->parent; while(inputNode && inputNode->key parent; } return inputNode; } } "See full answer

Gabriele G. - "What about exploiting the hash set and that is it? def smallestSubstring(s: str, t: str) -> str: if len(t) > len(s): return "" r = len(s) - 1 not_found = True while r > 0 and not_found: subs_set = set(s[0:r + 1]) for c in t: if not c in subs_set: not_found = False if not_found: r -= 1 else: r += 1 l = 0 not_found = True while l < r and not_"See full answer

Anonymous Armadillo - "I couldn't follow the solution offered here, but my solution seemed to pass 6/6 tests. Any feedback is welcome, thank you! def encrypt(word): en_word = "" for i in range(len(word)): if i == 0: en_word += chr(ord(word[0])+1) else: num = ord(word[i]) + ord(en_word[i-1]) while num > 122: num -= 26 en_word += chr(num) return en_word def decrypt(word): de_word = "" for i in range(len(word)): if i == 0: de_word += chr(ord(word[i]"See full answer

Anonymous Wasp - "I was able to provide the optimal approach and coded it up"See full answer

Anonymous Owl - "def getdifferentnumber(arr): \# arr of nonnegative ints \# find the smallest non negative int that isn't in the array MAX_INT = 2^31-1 \# not allowed to modify the input arr seen = [] smallest = 0 for i in arr: if smallest in arr: smallest += 1 seen.append(i) return smallest"See full answer

Anonymous Owl - "def getcheapestcost(rootNode): \# need to do DFS for each branch \# but this can be done recursively n = len(rootNode.children) if n == 0: return 0 else: min_cost = float('inf') for i in range(len(n)): tempcost = getcheapest_cost(rootNode.children[i]) if (tempcost < mincost): mincost = tempcost return min_cost + rootNode.cost \# A node class Node: \# Constructor to create a new node def init\(self, cost): self.cost = cost self.children = [] self.parent = None"See full answer

Alex M. - "A red-black tree is a self-balancing binary search tree. The motivation for this is that the benefits of O(logN) search, insertion, and deletion that a binary tree provides us will disappear if we let the tree get too "imbalanced" (e.g. there are too many nodes on one side of the tree or some branches have a depth that is way out of proportion to the average branch depth). This imbalance will occur if we don't adjust the tree after inserting or deleting nodes, hence our need for self-balancing c"See full answer