Recent Software Engineer Interview Questions

Ying T. - "def encode(root): if not root: return [] def dfs(node): if not node: return res.append(node.val) res.append(len(node,children)) for child_node in node.children: dfs(child_node) res = [] dfs(root) return res def decode(arr): if not arr: return None n = len(arr) i = 0 def dfs(val, children_count): if children_count == 0: return Node(val) cur_node = Node(val) cur_node.children = [] for j in range(children_count): nonlocal i i += 2 cur_node.children.append(dfs(arr[i], arr[i"See full answer

Khushbu R. - "Let’s say the matrix is m x n (i.e., m rows and n columns). Start from the top-right corner of the matrix. Move left if you see a 1. Move down if you see a 0. Keep track of the row index where you last saw the leftmost 1 — that row has the most 1s. public class MaxOnesRow { public static int rowWithMostOnes(int matrix) { int rows = matrix.length; int cols = matrix[0].length; int maxRowIndex = -1; int j = cols - 1; /"See full answer

Divyani .. - "Did the code in Python"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

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

Nesru M. - "Good to you The addition of an intermediate "sanitization" ML is something Neeraj used in the Uber Eats design and again seems kind of outside the scope here and redundant. This can simply be built into the AI response model to save a step. It's not clear what benefit this step provides,"See full answer

Javed S. - "One good reason for using the doubly linked list is that to remove an accessed item in the list from its existing position say in the middle of the list and move it to the head, you need to update the previous node's next pointer and the next node's previous pointer. A singly linked list only has next pointers, so removing from the middle would require traversing from the head to find the previous node which requires O(n) time. A doubly linked list allows O(1) removal from any position because w"See full answer