Apple Data Engineer Interview Questions

Anonymous Raccoon - "def find_primes(n): lst=[] for i in range(2,n+1): is_prime=1 for j in range(2,int(i**0.5)+1): if i%j==0: is_prime=0 break if is_prime: lst.append(i) return lst "See full answer

Prajwal M. - "from typing import List def maxprofitgreedy(stock_prices: List[int]) -> int: l=0 # buying r=1 # selling max_profit=0 while rstock_prices[l]: profit=stockprices[r]-stockprices[l] maxprofit=max(maxprofit,profit) else: l=r r+=1 return max_profit debug your code below print(maxprofitgreedy([7, 1, 5, 3, 6, 4])) `"See full answer

Anonymous Wasp - "I was able to answer this question and the follow-up questions as well"See full answer

Tiago R. - "const ops = { '+': (a, b) => a+b, '-': (a, b) => a-b, '/': (a, b) => a/b, '': (a, b) => ab, }; function calc(expr) { // Search for + or - for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if (['+', '-'].includes(char)) { return opschar), calc(expr.slice(i+1))); } } // Search for / or * for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if"See full answer

Komal S. - "a. Sort the array elements. b. take two pointers at index 0 and index Len-1; c. if the sum at the two pointers is target; break and return the pair d. if the sum is smaller, then move left pointer by 1 e. else move right pointer by 1; run the logic till the target is met or right pointer crosses the left pointer."See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

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

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"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

Tiago R. - "function permute(nums) { if (nums.length <= 1) { return [nums]; } const prevPermutations = permute(nums.slice(0, nums.length-1)); const currentNum = nums[nums.length-1]; const permutations = new Set(); for (let prev of prevPermutations) { for (let i=0; i < prev.length; i++) { permutations.add([...prev.slice(0, i), currentNum, ...prev.slice(i)]); } permutations.add([...prev, currentNum]); } return [...permutations]"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

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

Vince S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Kunal kumar S. - "simply check its size if the size if the size is greater than n then yes it has duplicate"See full answer

Tiago R. - "function isPalindrome(s, start, end) { while (s[start] === s[end] && end >= start) { start++; end--; } return end <= start; } function longestPalindromicSubstring(s) { let longestPalindrome = ''; for (let i=0; i < s.length; i++) { let j = s.length-1; while (s[i] !== s[j] && i <= j) { j--; } if (s[i] === s[j]) { if (isPalindrome(s, i, j)) { const validPalindrome = s.substring(i, j+1"See full answer