Meta Software Engineer Interview Questions

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Nikhil R. - "public class sample { public int [] merge(int [] a, int [] b) { if(a == null || a.length == 0 || b == null || b.length == 0) return null; int i = 0, j = 0, index = -1; int [] merged = new int[a.length + b.length]; while (i < a.length && j < b.length) { if(a[i] < b[i]) merged[++index] = a[i++]; else merged[++index] = b[j++]; } while (i < a.length) { merged[++index] = a[i++]; } "See full answer

Ds S. - "Initially I asked clarifying questions like whether the tree can be empty or not and asked the interviewer to explain what is meant by left view and the explanation for the sample inputs. Then I came up with the level order traversal approach where we visit each level in the binary tree at once using a queue and at each level print the value of the first node. Interviewer seemed satisfied with the approach and asked me to code it up. Finally gave the time and space complexity of the solution."See full answer

Srihitha J. - "my answer: void* memcpy(void* dest, const void* src, size_t n) { unsigned char* uDest = static_cast(dest); const unsigned char* ucSrc = static_cast(src); for(size_t i= 0; i(dest); const unsigned c"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Moshe S. - "Law is my passion. Traveling all over the world in 5 years"See full answer

Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

Anonymous Roadrunner - "from typing import List def traprainwater(height: List[int]) -> int: if not height: return 0 l, r = 0, len(height) - 1 leftMax, rightMax = height[l], height[r] res = 0 while l < r: if leftMax < rightMax: l += 1 leftMax = max(leftMax, height[l]) res += leftMax - height[l] else: r -= 1 rightMax = max(rightMax, height[r]) "See full answer

Tiago R. - "class TrieNode { constructor() { this.children = {}; this.isEndOfWord = false; } } class Trie { constructor() { this.root = new TrieNode(); } insert(word) { let node = this.root; for (const char of word) { if (!node.children[char]) { node.children[char] = new TrieNode(); } node = node.children[char]; } node.isEndOfWord = true; } search(word) { l"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