TikTok Data Scientist 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

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

Nikita B. - "Mission: Tiktok's mission is to inspire creativity and Joy. Any business wants to make sure that they are serving the value to their customers: For TikTok customers are: Viewers 2. Content Creators 3. Advertisers So few metrics we could measure are: Time spent/day Total no of videos created/day engagement rate = users who interacted in one of the meaningful action on Tiktok / total users at a day level either likes, share, watched vide for at least 5 mins, created video "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

Jeremy D. - "Any cycle would cause the prerequisite to be greater than the course. This passes all the tests: function canFinish(_numCourses, prerequisites) { for (const [a, b] of prerequisites) { if (b > a) return false } return true } `"See full answer

Devesh K. - "this solution here is much faster than the exponent reference soln. It is also far more concise and easy to understand def moveZerosToEnd(arr: List[int]) -> List[int]: left = 0 for right in range(len(arr)): if arr[right] == 0: pass else: if left != right: temp = arr[left] arr[left] = arr[right] arr[right] = temp left += 1 return arr `"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

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

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

Saipranay M. - "During the early learning stages of my career as a data scientist, I struggled to fully grasp the concept of patterns in seemingly random events. Despite thorough research, I remained skeptical about the universality of this idea. However, my perspective changed during a surprising scuba diving experience while on vacation. The underwater world I encountered not only amazed me but also gave me a key insight into patterns in nature and data analysis. As I dove into the clear waters, I witnessed"See full answer

Prajwal M. - "from typing import List def maxprofitgreedy(stock_prices: List[int]) -> int: l=0 # buying r=1 # selling max_profit=0 while rstock_prices[l]: profit=stockprices[r]-stockprices[l] maxprofit=max(maxprofit,profit) else: l=r r+=1 return max_profit debug your code below print(maxprofitgreedy([7, 1, 5, 3, 6, 4])) `"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

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Brook - "too much discussing on p-value…. and theoritical things…. country are independant…."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 Eagle - "I generate insights through stakeholder requirements and the data I have in hand"See full answer