Software Engineer Coding Interview Questions

Anni P. - "I firstly discuss the brute force approach in O(n^2) time complexity , than i moved to O(nlogn) tine complexity than i discussed the O(n) time complexity and O(n) space complexity . But interviewer want more optimised solution , in O(n) time complexity without using extra space , The solution wants O(1) space complexity i have to do changes in same array without using any space . This method is something like i have to place positive values to its original position by swapping and rest negativ"See full answer

Srikant V. - "Used Recursive approach to traverse the binary search tree and sum the values of the nodes that fall within the specified range [low, high]"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

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Anonymous Prawn - "Would be better to adjust resolution in the video player directly."See full answer

Karan K. - "2 Approaches: 1) The more intuitive approach is doing a multi-source BFS from all cats and storing the distance of closest cats. Then do a dfs/bfs from rat to bread. Time Complexity: O(mn + 4^L) where L is path length, worst case L could be mn Space Complexity: O(m*n) 2) The first approach should be fine for interviews. But if they ask to optimize it further, you can use Binary Search. Problems like "Finding max of min distance" or "Finding min of max" could be usually solved by BS. "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

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

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

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

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

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

J R. - "These are a set of utilities used to manage the heap memory as part of an application. The C standard library implements these functions. malloc(bytes) takes a number of bytes and returns a pointer to the start of the allocated buffer. If the allocation failed, a null pointer is returned instead. calloc(count, size) behaves like malloc(count * size), but also zero-initializes the allocated buffer, assuming the allocation succeeded. realloc(ptr, size) takes a pointer to a previously al"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 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