Goldman Sachs Interview Questions

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

Elle M. - "As a project or program manager to influence effectively without direct authority, focus on building credibility and trust by consistently delivering results, demonstrating empathy for stakeholder needs, and aligning everyone on shared goals. When managing up, tailor communication to leadership’s priorities and concerns—bring actionable insights, clear data, and potential solutions to the table. When managing down, empower teams by promoting ownership, offering timely recognition, and maintainin"See full answer

Manoj R. - "def traprainwater(height: List[int]) -> int: n = len(height) totalwaterlevel = 0 for i in range(n): j = i+1 while j = n: break rows = j - i -1 intrwaterlevel = min(height[j], height[i]) * rows for k in range(i+1, j): intrwaterlevel -= height[k] totalwaterlevel += intrwaterlevel i = j return totalwaterlevel"See full answer

Md kamrul H. - "public class HashMap { public class Element { T key; V value; Element(T k, V v) { this.key = k; this.value = v; } } private static final int DEFAULT_CAPACITY = 16; private static final float LOAD_FACTOR = 0.75f; private LinkedList[] table = new LinkedList[DEFAULT_CAPACITY]; private int size = 0; private int threshold = (int) (DEFAULTCAPACITY * LOADFACTOR); public void put(T k"See full answer

Ravi teja N. - "I solved it using recursion and then memoization. Used Dynamic programming approach"See full answer

Anonymous Wolf - "Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time. Maintain two zero-initialised variables - sum and count. Keep popping off the heap while sum < k, and update count. In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."See full answer

Tiago R. - "function isPalindrome(s, start, end) { while (s[start] === s[end] && end >= start) { start++; end--; } return end <= start; } function longestPalindromicSubstring(s) { let longestPalindrome = ''; for (let i=0; i < s.length; i++) { let j = s.length-1; while (s[i] !== s[j] && i <= j) { j--; } if (s[i] === s[j]) { if (isPalindrome(s, i, j)) { const validPalindrome = s.substring(i, j+1"See full answer

Shar N. - "standard answer for this."See full answer