Data Structures & Algorithms Interview Questions

Dadja Z. - "Traverse the array of points while computing the distance and pushing it to the heap. Then traverse again the heap and pop the k closest. Time O(nlogn) Space O(n)"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

XponentShift32 - "solving to find a cycle in directed graph"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

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

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

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

Kargi C. - "Average case - lookup/insert/delete - o(1) -> assuming a low load factor and uniform hash distribution. Worst case - o(n) -> where are keys collide in same bucket"See full answer

Bless M. - "You can ask some clarifying questions like 1) Ask if the list is already sorted or not 2) is zero included in the list ? 3) Natural numbers are usually positive numbers ( clarify they are non negatives) Solution : 1) If sorted use two pointers and sort them in O(N) 2) if not sorted , -ve / only +ve numbers in the list doesn't matter - the easiest solution is Use a priority queue and push the number and its square in each iteration Finally return the list returned by the priority Queue. N"See full answer

Vidhyadhar V. - "public class CircularBuffer { private T[] buffer; private int head; private int tail; private int size; private final int capacity; public CircularBuffer(int capacity) { this.capacity = capacity; this.buffer = (T[]) new Object[capacity]; this.head = 0; this.tail = 0; this.size = 0; } public void enqueue(T item) { if (isFull()) { throw new IllegalStateException("Buffer is full"); } buf"See full answer

Anzhe M. - "It's a 2Sum question with duplicate array elements."See full answer

Tiago R. - "function mostCommonWords(text) { const frequencyTable = new Map(); const words = text.toLowerCase().replace(//g, '').split(' '); for (let word of words) { frequencyTable.set(word, (frequencyTable.get(word) || 0)+1); } frequencyTable.delete(''); return [...frequencyTable.entries()].sort(([w1, f1], [w2, f2]) => f2-f1 !== 0? f2-f1 : w1.charCodeAt(0)-w2.charCodeAt(0) ); } `"See full answer

GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer

Rafał P. - "def changeString(org: str,target:str) -> bool: lOrg = len(org) lTarget = len(target) \# They have to be equal in lenght if lOrg != lTarget: return False counter1 = Counter(org) counter2 = Counter(target) \# Counter internally iterates through the input sequence, counts the number of times a given object occurs, and stores objects as keys and the counts as values. if counter1 != counter2: return False diff = sum(org[i] != target[i] for i in range(n)) return diff == 2 or (diff == 0 and any(v > 1 f"See full answer

Gabriele G. - "this assumes that the dependency among courses is in a growing order: 0 -> 1 -> 2 -> ... if not, then the code will not work"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

Anonymous Unicorn - "input_logs = [ f"{senderid} {receiverid} {transaction_count}" "1 2 2", "3 2 42", "2 2 22", "1 1 12", "2 1 1", "2 5 4", "4 2 15" ] input_threshold = 20 exptected_output = [ list of user_ids that made more than 20 transactions sorted by number of transactions in descending order "3", # 42 transactions "2", # 27 transactions (22 + 1 + 4) #"4", # 15 transactions #"1" # 14 transactions (2 + 12 + 1) ] def gettopapi_users(logs, thres"See full answer