Meta Coding Interview Questions

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

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

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

Anonymous Porcupine - "Coded the solution using this approach that is frequency table and counting. it is Leetcode 791"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

Ina K. - "HashMap supports insert, search, delete and retrieve in O(1). It stores data as key value pairs."See full answer

Rishabh R. - "if decreasing arr, start from end and keep checking if next element increases by 1 or not. wherever not, put that value there."See full answer

Rishabh N. - "I first asked few clarifying questions like the return array may need not contain the list of building in the same order, to which the interviewer agreed. Then I came up with an approach where we iterate the array from right to left and keep a max variable which will keep the value of the current max. When we find an item which is greater than max we update the max and add this element into our solution. The interviewer agreed for the approach. I discussed few corner scenarios with the interview"See full answer