Meta Coding Interview Questions

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

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

Michael B. - "Use an index, two pointers, and a set to keep track of elements that you've seen. pseudo code follows: for i, elem in enumerate(array): if elem in set return False if i > N: set.remove(array[i-N])"See full answer

Kun P. - "need to handle cases where src and dst overlaps: src starts before dst // start copying from end dst starts before src // start copying from beginning"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

Michael B. - "Problem: Given a modified binary tree, where each node also has a pointer to it's parent, find the first common ancestor of two nodes. Answer: As it happens, the structure that we're looking at is actually a linked list (one pointer up), so the problem is identical to trying to find if two linked lists share a common node. How this works is by stacking the two chains of nodes together so they're the same length. chain1 = node1 chain2= node2 while True: chain1 = chain1.next chain2=chain"See full answer

Rishabh R. - "sum of continuous subarray and keep checking if arr[i]==arr[j]. if true increase count;"See full answer

JOBHUNTER - "def findAlibaba(countOfRooms, strategy): #countofrooms: num rooms #listRooms rooms to look for alibabba possiblePlaces = [] #initialize rooms for i in range(countOfRooms): possiblePlaces.append(True) for i in range(len(strategy)): roomToCheck = strategy[i] #Room is marked as unavailable possiblePlaces[roomToCheck] = False #Next day calculatins nextDayPlaces = [] for j in range(countOfRooms): nextDayPla"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

Anonymous Porcupine - "https://www.geeksforgeeks.org/find-local-minima-array/ I coded O(N) but after that gave a binary approach aswell. After that he also gave a varient of this problem in which, local minima means that the number is strictly less than its adjacent (we cannot do binary search there sample test case [1,1,1,1,1,1,0,1] or [1,0,1,1,1,1,1,1] using mid we cannot determine if the minima is on left or right). we have to do a linear search or find recursively."See full answer