Adobe Machine Learning Engineer Interview Questions

Ugo C. - "function preorderToInorder(preorder) { let inorder = []; let stack = []; let root = preorder[0]; stack.push(root); for (let i = 1; i 0 && stack[stack.length - 1] 0) { root = stack.pop(); inorder.push(r"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

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

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

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

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

Reno 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

Areeba M. - "create a queue push all cells with 0 into queue Mark all 1s as unvisited (-1 or large value) Run BFS for each cell, explore 4 directions If neighbor is unvisited:distance = current + 1 push into queue "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

Relynn may silver B. - "import java.util.Arrays; import java.util.stream.Collectors; class Main { // Recursive function to print all combinations of numbers from \i\ to \n\ // having sum \n. The index\ denotes the next free slot in the output array \out\ public static void printCombinations(int i, int n, int[] out, int index) { // if the sum becomes n, print the combination if (n == 0) { System.out.println(Arrays.stream(out).limit(index) .boxed().collect(Collectors.toList())); } // start from the previous e"See full answer