Goldman Sachs Data Structures & Algorithms Interview Questions

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

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

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

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

Abinash S. - " def hasgoodsubarray(nums, k): if not nums: return False if k == 0: for i in range(len(nums)): if nums[i] == 0 and nums[i + 1] == 0: return True return False map = {0:-1} sum = 0 for i,val in enumerate(nums): sum += val rem = sum % k if rem in map: if i - map[rem] >= 2: return True else: map[rem] = i return False print(hasgoods"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

Jeff S. - " function isValid(s) { const openBrackets = new Set(['(','{','[']); const closeBrackets = new Map([ [')', '('], ['}', '{'], [']', '['] ]); const stack = []; for (let i = 0; i < s.length; i++) { if (openBrackets.has(s[i])) { stack.push(s[i]); } else { if (closeBrackets.get(s[i]) !== stack.pop()) { return false; } } } return stack.length === 0; } // deb"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

Maciej Z. - "function twoSum(nums, target) { const n = nums.length const map = new Map() for (let i=0; i<n; i++) { if (map.has(nums[i])) return [map.get(nums[i]), i] const diff = target - nums[i] map.set(diff, i) } return [] } `"See full answer

Laksitha R. - "static boolean hasCycle(ListNode head) { if (head == null || head.next == null) { return false; } ListNode slow = head; ListNode fast = head.next; while (slow != null && fast.next != null) { if (slow == fast) { return true; } slow = slow.next; fast = fast.next.next; } return false; } "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

Basil A. - " import java.io.*; import java.util.*; class Solution { static int trapRainWater(int[] height) { // your code goes here if(height.length == 0) { return 0; } int n = height.length; int[] left = new int[n]; int[] right = new int[n]; Arrays.fill(left, Integer.MAX_VALUE); Arrays.fill(right, Integer.MAX_VALUE); left[0] = height[0]; right[n-1] = height[n-1]; for(int i = 1; i < n - 1;"See full answer