Data Structures & Algorithms Interview Questions

Nils G. - "Bitshift the number to the right and keep track of the 1's you encounter. If you bitshift it completely and only encounter one 1, it is a power of two."See full answer

Tiago R. - "class TrieNode { constructor() { this.children = {}; this.isEndOfWord = false; } } class Trie { constructor() { this.root = new TrieNode(); } insert(word) { let node = this.root; for (const char of word) { if (!node.children[char]) { node.children[char] = new TrieNode(); } node = node.children[char]; } node.isEndOfWord = true; } search(word) { l"See full answer

Ahmed A. - "Good Question, but I would've marked this as medium not hard difficulty, since it's just a straightforward traversal."See full answer

Anonymous Wolf - "Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time. Maintain two zero-initialised variables - sum and count. Keep popping off the heap while sum < k, and update count. In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

Anonymous Owl - "def flatten_dictionary(dictionary): \# return a flattened dictionary - int/string/another dictionary values \# if the key is empty, exclude from the output \# concat using a "." btwn them \# add to res which is { "key.a.b.etc": "value" } \# iterate through the key value pairs \# while there is a key value pair in the value \# continue going through that, until the value is an int/string flatDic = {} flatDicHelper("", dictionary, flatDic) print(flatDic) return flatDic def flatDicHelper(initialKey"See full answer

Tiago R. - "function permute(nums) { if (nums.length <= 1) { return [nums]; } const prevPermutations = permute(nums.slice(0, nums.length-1)); const currentNum = nums[nums.length-1]; const permutations = new Set(); for (let prev of prevPermutations) { for (let i=0; i < prev.length; i++) { permutations.add([...prev.slice(0, i), currentNum, ...prev.slice(i)]); } permutations.add([...prev, currentNum]); } return [...permutations]"See full answer

Farhan -. - "Sorting is a technique to arrange data in either increasing order or decreasing order, and, the function that implements this functionality is called sort function"See full answer

Vishnu V. - "// Helper function to calculate the Euclidean distance between two points function distance(p1, p2) { return Math.sqrt(Math.pow(p1[0] - p2[0], 2) + Math.pow(p1[1] - p2[1], 2)); } // A helper function to find the closest pair in a given set of points within the strip function closestPairInStrip(strip, d) { let minDist = d; // Start with the current minimum distance strip.sort((a, b) => a[1] - b[1]); // Sort the strip by y-coordinate for (let i = 0; i < strip.length; i++) { "See full answer

Anonymous Hare - "function getDifferentNumber(arr) { // your code goes here const n = arr.length; //Define Max Integer const MAX_INT = Math.pow(2, 31) - 1; //Coppy arr to arr1 then sort arr1. const arr1 = arr; arr1.sort(function(a,b) {return a -b}); // Put arr1 in Set to optimize lo const uniqueSet = new Set(arr1); console.log(uniqueSet); // Check for the smallest nonnegative integer not in the array for (let i = 0; i < n; i++) { if (!uniqueSet.has(i)) { return i; } } if(n<MAX_INT) return n+1; else return -1; }"See full answer