Machine Learning Engineer Interview Questions

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

Deepak K. - "FN Given text need to figure out is it following guidelines. Should notify the user in case of not following guidelines. Reason for failure should have misleading/spam/adult filters. NFN High availability High Scalability Low latency of processing Estimations 1M requests/min text - 10kb => 9.5GB/min => 14TB/day API fetchmoderationscore(text) score will be between 0 to 1 more than 0.8 => not following guidelines fetchmoderationscore(text, filter)"See full answer

Anonymous Hare - "Sharing the approach for functional requirements we tool to solve this question. Functional Requirements This is only for the Registered users What is a "For You" page ? Home page where you get suggestions based on people you follow. Interactions like/share/comments (done by user) Interests (shared by the user during registration or onboarding) sports choices/ region choices/ Video sharing platform. So how many videos should we s"See full answer

Balaji G. - "My Answer. Please let me know your feedback. I have led many projects throughout my career. So, whenever I lead the project, I first try to clarify the goals and values delivered. For example, I was in charge of Data Analytics projects at my employment at a Startup. The team’s goal was launching the MVP. First, I discussed with the CEO to define the clear goal and scope for each of the projects. Next, I identified all the key stakeholders and received their inputs on the project's different a"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

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

Michael F. - "As an interviewer, I have asked this question to candidates in the past. Here are the major topics I am looking for in an interview The candidate should understand that there are ways of measuring the loss of a particular clustering. For example, we can take the average distance of each point to it's cluster center. The candidate should understand that this loss will always decrease as the number of clusters increases. For that reason, we can't just pick the value of K that minimizes the l"See full answer

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

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