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Machine Learning Engineer Coding Interview Questions

Review this list of 80 Coding Machine Learning Engineer interview questions and answers verified by hiring managers and candidates.
  • Adobe logoAsked at Adobe 
    Video answer for 'Edit distance'
    +32

    "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)): "

    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

    Machine Learning Engineer
    Coding
    +3 more
  • Adobe logoAsked at Adobe 
    +36

    "Was this for an entry level engineer role?"

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

    Machine Learning Engineer
    Coding
    +4 more
  • Adobe logoAsked at Adobe 
    +24

    "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"

    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

    Machine Learning Engineer
    Coding
    +6 more
  • +1

    " 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"

    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

    Machine Learning Engineer
    Coding
    +4 more
  • Apple logoAsked at Apple 
    +26

    "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"

    Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

    Machine Learning Engineer
    Coding
    +4 more
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  • +11

    "Would be better to adjust resolution in the video player directly."

    Anonymous Prawn - "Would be better to adjust resolution in the video player directly."See full answer

    Machine Learning Engineer
    Coding
    +4 more
  • Anthropic logoAsked at Anthropic 
    Machine Learning Engineer
    Coding
    +4 more
  • Adobe logoAsked at Adobe 

    "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."

    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

    Machine Learning Engineer
    Coding
    +4 more
  • Adobe logoAsked at Adobe 
    +26

    "#inplace reversal without inbuilt functions def reverseString(s): chars = list(s) l, r = 0, len(s)-1 while l < r: chars[l],chars[r] = chars[r],chars[l] l += 1 r -= 1 reversed = "".join(chars) return reversed "

    Anonymous Possum - "#inplace reversal without inbuilt functions def reverseString(s): chars = list(s) l, r = 0, len(s)-1 while l < r: chars[l],chars[r] = chars[r],chars[l] l += 1 r -= 1 reversed = "".join(chars) return reversed "See full answer

    Machine Learning Engineer
    Coding
    +4 more
  • Google logoAsked at Google 
    Video answer for 'Merge Intervals'
    +45

    "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"

    Kofi N. - "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"See full answer

    Machine Learning Engineer
    Coding
    +6 more
  • +2

    "Started off with brute force then optimized solution"

    Nikhil R. - "Started off with brute force then optimized solution"See full answer

    Machine Learning Engineer
    Coding
    +2 more
  • Apple logoAsked at Apple 
    +20

    "def is_valid(s: str) -> bool: stack = [] closeToOpen = { ")" : "(", "]" : "[", "}" : "{" } for c in s: if c in closeToOpen: if stack and stack[-1] == closeToOpen[c]: stack.pop() else: return False else: stack.append(c) return True if not stack else False debug your code below print(is_valid("()[]")) `"

    Anonymous Roadrunner - "def is_valid(s: str) -> bool: stack = [] closeToOpen = { ")" : "(", "]" : "[", "}" : "{" } for c in s: if c in closeToOpen: if stack and stack[-1] == closeToOpen[c]: stack.pop() else: return False else: stack.append(c) return True if not stack else False debug your code below print(is_valid("()[]")) `"See full answer

    Machine Learning Engineer
    Coding
    +4 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Move all zeros to the end of an array.'
    +59

    "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. "

    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

    Machine Learning Engineer
    Coding
    +4 more
  • Amazon logoAsked at Amazon 
    +5

    "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "

    Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer

    Machine Learning Engineer
    Coding
    +4 more
  • Meta logoAsked at Meta 
    +2

    "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."

    דניאל ר. - "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."See full answer

    Machine Learning Engineer
    Coding
    +2 more
  • Amazon logoAsked at Amazon 
    +3

    "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"

    Khushbu R. - "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"See full answer

    Machine Learning Engineer
    Coding
    +2 more
  • +13

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

    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

    Machine Learning Engineer
    Coding
    +4 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Product of Array Except Self'
    +58

    "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?"

    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

    Machine Learning Engineer
    Coding
    +3 more
  • Adobe logoAsked at Adobe 
    +29

    " 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])) `"

    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

    Machine Learning Engineer
    Coding
    +2 more
  • Amazon logoAsked at Amazon 
    Video answer for 'Implement k-means clustering.'

    "at first I want to know number of cluster I will put random number if I don't know and I will use method called Elbow method or Silhouette Score ,Gap Statistic and Davies–Bouldin Index to know the best number of cluster and I will use scikit-learn library to import kmeans from sklearn.cluster import KMeans kmeans = KMeans(nclusters=2, randomstate=0) kmeans.fit(X) and X this my data "

    Taheia S. - "at first I want to know number of cluster I will put random number if I don't know and I will use method called Elbow method or Silhouette Score ,Gap Statistic and Davies–Bouldin Index to know the best number of cluster and I will use scikit-learn library to import kmeans from sklearn.cluster import KMeans kmeans = KMeans(nclusters=2, randomstate=0) kmeans.fit(X) and X this my data "See full answer

    Machine Learning Engineer
    Coding
    +5 more
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