Data Structures & Algorithms Interview Questions

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Rahul M. - "Idea for solution: Reverse the complete char array Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters. vector reverseSubarray(vector& arr, int s, int e) { while (s reverseWords(vector& arr ) { int n = arr.size(); reverse(arr, 0, n - 1"See full answer

Kofi N. - "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"See full answer

Ramachandra N. - "from collections import deque from typing import List def longestsubarraydifflessthan_n(nums: List[int], N: int) -> int: """ Find the length of the longest contiguous subarray such that the difference between any two elements in the subarray is less than N. Equivalent condition: max(subarray) - min(subarray) < N Approach (Optimal): Sliding window with two monotonic deques: max_d: decreasing deque of indices (front is index of current max"See full answer

Frederick K. - "\# An program that prints out the peak elements in a list of integers. Pseudocode: 1. Define a function that takes a list of integers as input. 2. Initialize an empty list to store the peak elements. 3. Loop through the list of integers. 4. For each element, check if it is greater than its neighbors. 5. If it is, add it to the list of peak elements. 6. Return the list of peak elements. def findpeakelements(nums): if not nums: return [] peaks = [] n = len(nums"See full answer

Venkata rakesh M. - "Here if we breakdown each dependency [A,B] , We need to check if there a cycle in Dependency Graph. If there is cycle installation is not possible, If there is no cycle installation is possible. Steps : 1: Build the graph 2: Perform DFS based Cycle Detection 3: Check each package if those packages have cycle or not."See full answer

Nicholas S. - "First thing the interviewee did wrong is not asking clarifying questions. This is the most vague problem I have every heard, and the interviewee just made assumptions and started programming."See full answer

Yugaank K. - "A recursive backtracking solution in python. def changeSigns(nums: List[int], S: int) -> int: res = [] n = len(nums) def backtrack(index, curr, arr): if curr == S and len(arr) == n: res.append(arr[:]) return if index >= len(nums): return for i in range(index, n): add +ve number arr.append(nums[i]) backtrack(i+1, curr + nums[i], arr) arr.pop() "See full answer

Aashka C. - "Sorted the array and stored the minimum difference in a variable and then traversed the array for the pairs having minimum difference"See full answer

Gaurav B. - "class TreeNode(var val: Int, var left: TreeNode? = null, var right: TreeNode? = null) fun isAverageOfDescendants(root: TreeNode?): Boolean { fun helper(node: TreeNode?): Triple { if (node == null) return Triple(0, 0, true) val (leftSum, leftCount, leftValid) = helper(node.left) val (rightSum, rightCount, rightValid) = helper(node.right) val totalSum = leftSum + rightSum val totalCount = leftCount + rightCount // If leaf n"See full answer

Firdous A. - "Approach (BFS + Horizontal Distance) Assign a horizontal distance (HD) to each node. Root → HD = 0 Left child → HD = parent HD - 1 Right child → HD = parent HD + 1 Do a BFS (level order traversal). If a node with a given HD is seen for the first time, add it to the result. Ignore later nodes with the same HD (because only the top one is visible). After traversal, sort by HD and print nodes left to righ"See full answer

Rahul J. - "Constraints: 4-direction moves; no mode switching (pick exactly one of {1=bicycle, 2=bike, 3=car, 4=bus} for the full trip). Per-mode search: If a mode’s per-step time/cost are uniform, run BFS on allowed cells. Then totaltime = steps × timeperstep, tie-break by steps × costper_step. If time/cost vary by cell (given matrices), run Dijkstra per mode minimizing (totaltime, totalcost) lexicographically. Maintain the best ⟨time, cost⟩ per cell; relax when the new pair is strictly better. S"See full answer

Anonymous Goat - "Batch Packing Problem In Amazon’s massive warehouse inventory, there are different types of products. You are given an array products of size n, where products[i] represents the number of items of product type i. These products need to be packed into batches for shipping. The batch packing must adhere to the following conditions: No two items in the same batch can be of the same product type. The number of items packed in the current batch must be strictly greater than the number pack"See full answer

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

Ankush G. - " max Min 4, 3, 1 , 6, 7, 8 1 3 4 6 7 8 9 0 1 2 3 0 1 1 2 3 6 7 8 9 class MedianFinder{ std::priority_queue minHeap; std::priority_queue, greater> maxHeap; int numEleMaxheap = 0, numEleMinHeap = 0; public: void addNum( int n) { if(numEleMaxheap == numEleMinHeap ) { maxHeap.push(n); int maxofmaxheap = maxHeap.top(); maxHeap.pop(); "See full answer

Sh R. - "i responded using a multi sourced BFS and in place marking, then i checked the final grid to see if any free spots were left unmarked."See full answer

Kanishvaran P. - "class Solution { public boolean isValid(String s) { // Time Complexity and Space complexity will be O(n) Stack stack=new Stack(); for(char c:s.toCharArray()){ if(c=='('){ stack.push(')'); } else if(c=='{'){ stack.push('}'); } else if(c=='['){ stack.push(']'); } else if(stack.pop()!=c){ return false; } } return stack.isEmpty(); } }"See full answer

Ramachandra N. - " def closest_palindrome(n: str) -> str: """ Finds the closest palindromic number to n (excluding itself). Assumptions: If two palindromes are equally close, return the smaller one. n is a positive integer represented as a string. Time Complexity: O(1) Space Complexity: O(1) """ length = len(n) num = int(n) Helper to build palindrome from a prefix def makepalindrome(prefix: int, isodd_length: bool) -> int: s = str(prefi"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer