LinkedIn 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

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

Will I. - "Is it bad to get the answer a different way? Will they mark that as not knowing Bayes Theorem or just correct as it is an easier way to get the answer? The way I went is to look at what happens when the factory makes 100 light bulbs. Machine A makes 60 of which 3 are faulty, Machine B makes 40 of which 1.2 are faulty. Therefore the pool of faulty lightbulbs is 3/4.2 = 5/7 from machine A and 1.2/4.2 = 3/7 from Machine B."See full answer

Dinesh K. - "i dont know"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

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

Samuel M. - "function main(){ const v1=[2,3, 4, 10] const v2= [3,4 ,5,20, 23] return merge(v1,v2); } function merge(left, right){ const result=[]; while(left.length>0&& right.length>0){ if(left[0]0){ result=result.concat(left) } if(right.length>0){ result=result.concat(right) } return result; }"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

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

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer

Iris F. - "#include #include #include using namespace std; void printComs(int prev, int start, int end, int target) { if (start >= end) return; while (start target) { end--; } else { st"See full answer