Apple Interview Questions

Anonymous Jaguar - "Code reviews, automated linting, unit tests, integration tests, validation tests"See full answer

Medha A. - "Adoption- The number of successful calls by users per day or week. Engagement- Average time duration of the FaceTime call."See full answer

Musonda C. - "Imagine you have a friend from France that speaks French, and you only speak English. You and your friend want to play a talking game. So, for your friend to understand you and for you to understand what your friend is saying you would need an API between the two of you. The api has its own language called api language. You and your friend will both get a small book that tells you how to speak api language. The api in between you and your friend would allow both of you to speak API language to"See full answer

Hari priya K. - "Hey Grandma, you've had a lot of experience with infants, haven't you? When they were babies, you taught them how to chew in their first six months. This initial phase is like giving them data. Once they learned how to chew, they could handle any food you gave them. Next, you refined their learning by teaching them that they should only chew on food. This is like refining the data so they understand what is relevant. Then, a few months later, they started crawling and walking, learning by observ"See full answer

Abhilash K. - "Minimum viable product is a product with a minimum set of features that are sufficient to gather information if customers are going to use it or are sufficient to answer the questions for the problems that are we are specifically trying to solve. In order to prioritize features it is critical to consider following points. What is the problem we are trying to solve? Understand the context. Identify the features that you need to solve the problem while ensuring that features are in sync with"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

Abdurhman M. - "program management is a role similar to project management but often encompasses more that just managing projects."See full answer

Anonymous Wasp - "I was able to provide the optimal approach and coded it up"See full answer

Alex M. - "A red-black tree is a self-balancing binary search tree. The motivation for this is that the benefits of O(logN) search, insertion, and deletion that a binary tree provides us will disappear if we let the tree get too "imbalanced" (e.g. there are too many nodes on one side of the tree or some branches have a depth that is way out of proportion to the average branch depth). This imbalance will occur if we don't adjust the tree after inserting or deleting nodes, hence our need for self-balancing c"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

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

Divya R. - "static boolean sudokuSolve(char board) { return sudokuSolve(board, 0, 0); } static boolean sudokuSolve(char board, int r, int c) { if(c>=board[0].length) { r=r+1; c=0; } if(r>=board.length) return true; if(boardr=='.') { for(int num=1; num<=9; num++) { boardr=(char)('0' + num); if(isValidPosition(board, r, c)) { if(sudokuSolve(board, r, c+1)) return true; } boardr='.'; } } else { return sudokuSolve(board, r, c+1); } return false; } static boolean isValidPosition(char b"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