TikTok Software Engineer Interview Questions

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Rajesh V. - "As a PM i received a feedback from my program manager on my style of verbal communication. It is about me speaking faster when i wanted to get away with a topic that i wasn't confident (may be not backed up with data, or still in process of getting detailed insight of a problem etc.). Whereas when I'm confident I tend to speak slowly or more assertively that made people to follow easily. I welcomed that feedback so from then on when I'm not confident in a topic I became more assertive to let pe"See full answer

Pramod V. - "Referring to https://www.forbes.com/sites/forbesbusinesscouncil/2022/03/23/15-strategies-for-balancing-competing-stakeholder-priorities/?sh=7c82aa68262f Understand the conflicting priorities and align it with the goal/ objectives and the company mission. Start with the Least Common Denominator between the conflicting priorities to come to a commonality and start from there to objectively approach the next imp priority Always keep communication on and be transparent with 'equality' an"See full answer

Anubhav S. - " At low level: I would use two stacks: one for forward history and other for backward history. i go to tryexponent.com => this url will be stored in backward history stack. i go to google => again this url will be stored in backward history stack. i press back => data from backward history will be popped and put in to forward history stack. I press forward => data from forward history stack will be popped and put in to backward history tab. Also, whenever i go to any url,"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

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

Aymen D. - "It would have been more interesting to focus on the system design rather than the Trie DS, Interviewee could have just mentioned the Trie and passed to things more important. Interviewee should have focused on the factors on which he wants to scale the API servers, popularity of the query parts ? region may be ? A hash of many factors ? Caches should have definitely be discussed, Cache eviction policies, Cache invalidation managements... Interviewee should have mentioned which kind of API pro"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

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

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

V V. - "Simple file system ( Hash Map , LSM trees), and consistent hashing. I failed to discuss about conflicts ( Versioning)"See full answer

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

Divya R. - "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Tiago R. - "function findPrimes(n) { if (n < 2) return []; const primes = []; for (let i=2; i <= n; i++) { const half = Math.floor(i/2); let isPrime = true; for (let prime of primes) { if (i % prime === 0) { isPrime = false; break; } } if (isPrime) { primes.push(i); } } return primes; } `"See full answer

Jerry O. - "\# Definition for a binary tree node. class TreeNode: def init(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def maxPathSum(self, root: TreeNode) -> int: self.max_sum = float('-inf')"See full answer

Bravin V. - "I aim to utilize hands-on activities to apply the knowledge I've acquired, safeguard the organization from threat actors, and contribute to its growth."See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer