Adobe Machine Learning Engineer Interview Questions

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Yeshwanth D. - "Was this for an entry level engineer role?"See full answer

Ashley C. - "I follow a variation of the RICE framework when prioritizing how I ship product features. I start by looking at: Reach: Because the customer segmentation across our product portfolio is so similar, I tend to hold a lot of weight on product features that will maximize our customer reach with a minimal LOE. Impact: After establishing which customer segments will benefit from the product feature, I determine the urgency and estimated impact on each customer segment based on customer i"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Abinash S. - " def hasgoodsubarray(nums, k): if not nums: return False if k == 0: for i in range(len(nums)): if nums[i] == 0 and nums[i + 1] == 0: return True return False map = {0:-1} sum = 0 for i,val in enumerate(nums): sum += val rem = sum % k if rem in map: if i - map[rem] >= 2: return True else: map[rem] = i return False print(hasgoods"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Nicolás N. - "Two pointers sub-routine from typing import List def reverse_words(arr: List[str]) -> List[str]: size = len(arr) #reverse whole array reversesubarr(arr, 0, size-1) #reverse each word start, end = 0, 0 for i in range(size): if arr[i].isspace() or i == size-1: end = i-1 if i != size-1 else i reversesubarr(arr, start, end) start = i+1 return arr def reversesubarr(array, start, end): print(array, start,"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Laksitha R. - "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Rick E. - " from typing import List one pass O(n) def find_duplicates(arr1: List[int], arr2: List[int]) -> List[int]: duplicates = [] i1 = i2 = 0 while i1 < len(arr1) and i2 < len(arr2): if arr1[i1] == arr2[i2]: duplicates.append(arr1[i1]) i2 += 1 i1 += 1 return duplicates debug your code below print(find_duplicates([1, 2, 3, 5, 6, 7], [3, 6, 7, 8, 20])) `"See full answer

Jean-pierre C. - "function twoSum(nums, target) { let complements = new Map(); for (let i = 0; i < nums.length; i++) { let diff = target - nums[i]; if (complements.has(diff)) { return [complements.get(diff), i]; } complements.set(nums[i], i); } return []; } console.log(twoSum([2, 7, 11, 15], 9)); `"See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

Victor O. - " import java.util.*; class Solution { // Time Complexity: O(n^2) // Space Complexity: O(n) public static List> threeSum(int[] nums) { // Ensure that the array is sorted first Arrays.sort(nums); // Create the results list to return List> results = new ArrayList(); // Iterate over the length of nums for (int i = 0; i < nums.length-2; i++) { // We will have the first number in"See full answer

Tiago R. - "function findPrimes(n) { if (n < 2) return []; const primes = []; for (let i=2; i <= n; i++) { const half = Math.floor(i/2); let isPrime = true; for (let prime of primes) { if (i % prime === 0) { isPrime = false; break; } } if (isPrime) { primes.push(i); } } return primes; } `"See full answer

Rick E. - " O(n) time from typing import List def generate_parentheses(n: int): res = [] def generate(buf, opened, closed): if len(buf) == 2 * n: if n != 0: res.append(buf) return if opened < n: generate( buf + "(", opened + 1, closed) if closed < opened: generate(buf + ")", opened, closed + 1) generate("", 0, 0) return res debug your code below print(generate_parentheses(1"See full answer

Jerry O. - "\# Definition for a binary tree node. class TreeNode: def init(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def maxPathSum(self, root: TreeNode) -> int: self.max_sum = float('-inf')"See full answer

Sarvesh G. - "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"See full answer