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Machine Learning Engineer Data Structures & Algorithms Interview Questions

Review this list of 74 Data Structures & Algorithms Machine Learning Engineer interview questions and answers verified by hiring managers and candidates.
  • Google logoAsked at Google 
    4 answers
    +1

    "You can ask some clarifying questions like 1) Ask if the list is already sorted or not 2) is zero included in the list ? 3) Natural numbers are usually positive numbers ( clarify they are non negatives) Solution : 1) If sorted use two pointers and sort them in O(N) 2) if not sorted , -ve / only +ve numbers in the list doesn't matter - the easiest solution is Use a priority queue and push the number and its square in each iteration Finally return the list returned by the priority Queue. N"

    Bless M. - "You can ask some clarifying questions like 1) Ask if the list is already sorted or not 2) is zero included in the list ? 3) Natural numbers are usually positive numbers ( clarify they are non negatives) Solution : 1) If sorted use two pointers and sort them in O(N) 2) if not sorted , -ve / only +ve numbers in the list doesn't matter - the easiest solution is Use a priority queue and push the number and its square in each iteration Finally return the list returned by the priority Queue. N"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +1 more
  • Adobe logoAsked at Adobe 
    53 answers
    +49

    " Brute Force Two Pointer Solution: from typing import List def two_sum(nums, target): for i in range(len(nums)): for j in range(i+1, len(nums)): if nums[i]+nums[j]==target: return [i,j] return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"

    Ritaban M. - " Brute Force Two Pointer Solution: from typing import List def two_sum(nums, target): for i in range(len(nums)): for j in range(i+1, len(nums)): if nums[i]+nums[j]==target: return [i,j] return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +5 more
  • Accenture logoAsked at Accenture 
    16 answers
    +11

    "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"

    Divya R. - "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +3 more
  • Apple logoAsked at Apple 
    2 answers

    "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently. Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"

    Stanley Y. - "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently. Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +2 more
  • Adobe logoAsked at Adobe 
    Add answer
    Video answer for 'Find the median of two sorted arrays.'
    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
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  • Adobe logoAsked at Adobe 
    18 answers
    Video answer for 'Given an nxn grid of 1s and 0s, return the number of islands in the input.'
    +15

    " 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) "

    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

    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    23 answers
    Video answer for 'Find a triplet in an array with a given sum.'
    +17

    "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "

    Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +3 more
  • Apple logoAsked at Apple 
    17 answers
    +12

    "class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return True return False debug your code below node1 = ListNode(1) node2 = ListNode(2) node3 = ListNode(3) node4 = ListNode(4) creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"

    Anonymous Roadrunner - "class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return True return False debug your code below node1 = ListNode(1) node2 = ListNode(2) node3 = ListNode(3) node4 = ListNode(4) creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    12 answers
    +8

    "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; } `"

    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

    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • Airbnb logoAsked at Airbnb 
    Add answer
    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    12 answers
    Video answer for 'Generate Parentheses'
    +8

    " 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"

    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

    Machine Learning Engineer
    Data Structures & Algorithms
    +3 more
  • Adobe logoAsked at Adobe 
    2 answers
    Video answer for 'Given the root of a binary tree of integers, return the maximum path sum.'

    "\# 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')"

    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

    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • "Write a function which Caesar ciphers all the strings so that the first character is "a". Use ascii code points and the modulo operator to do this. Use this function to create a hashmap between each string and the CC-a string. Then go through each key:value pair in the hashmap, and use the CC-a ciphered value as the key in a new defaultdict(list), adding the original string to the value field in the output."

    Michael B. - "Write a function which Caesar ciphers all the strings so that the first character is "a". Use ascii code points and the modulo operator to do this. Use this function to create a hashmap between each string and the CC-a string. Then go through each key:value pair in the hashmap, and use the CC-a ciphered value as the key in a new defaultdict(list), adding the original string to the value field in the output."See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +2 more
  • Adobe logoAsked at Adobe 
    5 answers
    +2

    "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"

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

    Machine Learning Engineer
    Data Structures & Algorithms
    +3 more
  • Adobe logoAsked at Adobe 
    10 answers
    +7

    "function addChildren(root, val, inorder) { const rootInOrderIndex = inorder.indexOf(root.value); const childrenInOrderIndex = inorder.indexOf(val); if (childrenInOrderIndex < rootInOrderIndex) { if (!root.left) { root.left = new TreeNode(val); } else { addChildren(root.left, val, inorder); } } else { if (!root.right) { root.right = new TreeNode(val); } else { addChildren(root.right,"

    Tiago R. - "function addChildren(root, val, inorder) { const rootInOrderIndex = inorder.indexOf(root.value); const childrenInOrderIndex = inorder.indexOf(val); if (childrenInOrderIndex < rootInOrderIndex) { if (!root.left) { root.left = new TreeNode(val); } else { addChildren(root.left, val, inorder); } } else { if (!root.right) { root.right = new TreeNode(val); } else { addChildren(root.right,"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +2 more
  • Google logoAsked at Google 
    9 answers
    +6

    " from typing import List def least_interval(tasks: List[str], n: int) -> int: pass # your code goes here if n == 0: return len(tasks) dictionary = {} total_sum = len(tasks) output = 0 for t in tasks: if t in dictionary: dictionary[t] += 1 else: dictionary[t] = 1 dictLen = len(dictionary) while total_sum > 0: for key in dictionary.keys(): if dictionary[key] > 0: "

    Anonymous Quail - " from typing import List def least_interval(tasks: List[str], n: int) -> int: pass # your code goes here if n == 0: return len(tasks) dictionary = {} total_sum = len(tasks) output = 0 for t in tasks: if t in dictionary: dictionary[t] += 1 else: dictionary[t] = 1 dictLen = len(dictionary) while total_sum > 0: for key in dictionary.keys(): if dictionary[key] > 0: "See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +2 more
  • Meta logoAsked at Meta 
    4 answers
    +1

    "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"

    GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +2 more
  • Adobe logoAsked at Adobe 
    30 answers
    +22

    "This problem could be solved in two ways(both using Kadane's algorithm): Simple iterating 1-D dp function maxSubarraySum(nums) { const n = nums.length; if ( n === 0) return 0; const dp = Array(n).fill(0); dp[0] = nums[0]; for (let i = 1; i < n; i++) { dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]); } return Math.max(...dp); } "

    Mark K. - "This problem could be solved in two ways(both using Kadane's algorithm): Simple iterating 1-D dp function maxSubarraySum(nums) { const n = nums.length; if ( n === 0) return 0; const dp = Array(n).fill(0); dp[0] = nums[0]; for (let i = 1; i < n; i++) { dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]); } return Math.max(...dp); } "See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    13 answers
    Video answer for 'Merge k sorted linked lists.'
    +7

    "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"

    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

    Machine Learning Engineer
    Data Structures & Algorithms
    +4 more
  • Meta logoAsked at Meta 
    1 answer

    "Problem: Given an input string txt consisting of alphanumeric characters and the parentheses characters '(' & ')', write a function which removes the minimum number of characters to return a version of the string with properly balanced parenthesis. Answer: You can do this with a counter. Psuedo-Python Start with counter = 0 output = [] Iterate through the string, every time you encounter a '(', increment the counter. Add the character to the output. If you encounter a ')', decrement the coun"

    Michael B. - "Problem: Given an input string txt consisting of alphanumeric characters and the parentheses characters '(' & ')', write a function which removes the minimum number of characters to return a version of the string with properly balanced parenthesis. Answer: You can do this with a counter. Psuedo-Python Start with counter = 0 output = [] Iterate through the string, every time you encounter a '(', increment the counter. Add the character to the output. If you encounter a ')', decrement the coun"See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +1 more
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