Machine Learning Engineer Coding Interview Questions

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

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

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

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

Jeff S. - " class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function diameterOfTree(root) { if (root === null || root.left === null & root.right === null) { return 0; } function countBranch(node, count) { if (node.left === null && node.right === null) { return count; } let left = node.left === null ? 0 : countBranch(node.left, count+1); let right = no"See full answer

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"See full answer

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

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

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

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

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

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

Cagdas A. - " class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: pass # your code goes here if head is None: return False previousNodes = set() iter = head while iter: if iter.val in previousNodes: return True previousNodes.add(iter.val) iter = iter.next; return False debug your code below node1 = ListNode(1) node2 = ListNode(2) n"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

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