Data Scientist Data Structures & Algorithms Interview Questions

Richard W. - "int getMaxWater(vector& nums) { int n = nums.size(); int mx = INT_MIN; int i=0, j=n-1; while(i<j) { int water = (j - i) * min(nums[i], nums[j]); mx = max(mx, water); if(nums[i] < nums[j]){ i++; } else { j--; } } return mx; } `"See full answer

Satyam S. - "Reversing a linked list is a very popular question. We have two approaches to reverse the linked list: Iterative approach and recursion approach. Iterative approach (JavaScript) function reverseLL(head){ if(head === null) return head; let prv = null; let next = null; let cur = head; while(cur){ next = cur.next; //backup cur.next = prv; prv = cur; cur = next; } head = prv; return head; } Recursion Approach (JS) function reverseLLByRecursion("See full answer

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

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

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

Anonymous Prawn - "Would be better to adjust resolution in the video player directly."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

Anonymous Possum - "#inplace reversal without inbuilt functions def reverseString(s): chars = list(s) l, r = 0, len(s)-1 while l < r: chars[l],chars[r] = chars[r],chars[l] l += 1 r -= 1 reversed = "".join(chars) return reversed "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

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

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

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

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

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

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

Anonymous Roadrunner - "class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return True return False debug your code below node1 = ListNode(1) node2 = ListNode(2) node3 = ListNode(3) node4 = ListNode(4) creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"See full answer