Data Structures & Algorithms Interview Questions

Laksitha R. - "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "See full answer

Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Sailaja R. - "import java.util.*; public class MostCommonWords { public static String mostCommonWords(String text) { String removeNonAlphabets = "[.,;!\"'\\(\\)]"; String sanitizedText = text.replaceAll(removeNonAlphabets," ").toLowerCase(); String[] wordsArray = sanitizedText.split("\\s+"); Map hm = new HashMap(); for (String word: wordsArray){ if (!word.isEmpty()){ hm.put(word, hm.getOrDefault(word,0)+1); } } List> entries = new ArrayList(hm.entrySet()); en"See full answer

Khushbu R. - "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"See full answer

דניאל ר. - "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."See full answer

Sohum S. - "I would assume that this is similar to an intervals question. Meeting Rooms II (https://www.lintcode.com/problem/919/?fromId=203&_from=collection) on Leetcode seems like the closest comparison, it's a premium question so I linked Lintcode. I'm assuming that we also need to just return the minimum number of cars used. You need to sort for the most optimal solution, so you're constrained by an O(nlogn) time complexity. So any sorting solution could work (using a heap, sorting the array input arra"See full answer

Victor H. - "There is a faster approach that solves the problem in O(n) time: def find_duplicates(arr1, arr2): arr1 = set(arr1) res = [] for num in arr2: if num in arr1: res.append(num) return res `"See full answer

Shikha S. - "Binary serach on E range"See full answer

Praveen D. - "Let me try to explain it with simple life analogy You're cooking dinner in the kitchen. Multithreading is when you've got a bunch of friends helping out. Each friend does a different job—like one chops veggies while another stirs a sauce. Everyone focuses on their task, and together, you all make the meal faster. In a computer, it's like different jobs happening all at once, making stuff happen quicker, just like having lots of friends helping makes dinner ready faster."See full answer

Tiago R. - "function serialize(list) { for (let i=0; i 0xFFFF) { throw new Exception(String ${list[i]} is too long!); } const prefix = String.fromCharCode(length); list[i] = ${prefix}${list[i]}; console.log(list[i]) } return list.join(''); } function deserialize(s) { let i=0; const length = s.length; const output = []; while (i < length) { "See full answer

Jeff S. - " class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function diameterOfTree(root) { if (root === null || root.left === null & root.right === null) { return 0; } function countBranch(node, count) { if (node.left === null && node.right === null) { return count; } let left = node.left === null ? 0 : countBranch(node.left, count+1); let right = no"See full answer

Krishnaveni G. - "Since a bitonic array first increases then decreases, we can: Find the peak using binary search (O(log n)) Reverse the decreasing half Merge the two sorted halvesThis gives an overall time complexity of O(n)."See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Batman X. - "def friend_distance(friends, userA, userB): step = 0 total_neighs = set() llen = len(total_neighs) total_neighs.add(userB) while len(total_neighs)!=llen: s = set() step += 1 llen = len(total_neighs) for el in total_neighs: nes = neighbours(friends, userA, el) if userA in nes: return step for p in nes: s.add(p) for el in s: total_neighs.add(el) return -1 def neighbours(A,n1, n2): out = set() for i in range(len(A[n2])): if An2: out.add(i) return out"See full answer

Chris R. - "It was like say we have a library A which has a library B as a dependency and so on, how would we determine in the dependency chain that whether there is a circular depedency?"See full answer