Machine Learning Engineer Coding Interview Questions

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

Yash N. - "import time class Task: def init\(self, description, interval=None): self.description = description self.interval = interval self.next_run = time.time() class SimpleTaskScheduler: def init\(self): self.tasks = [] def add_task(self, description, interval=None): self.tasks.append(Task(description, interval)) def run(self, duration=60): end_time = time.time() + duration while time.time() < end_time: curr"See full answer

Ugo C. - "function preorderToInorder(preorder) { let inorder = []; let stack = []; let root = preorder[0]; stack.push(root); for (let i = 1; i 0 && stack[stack.length - 1] 0) { root = stack.pop(); inorder.push(r"See full answer

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

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

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

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

Anonymous Condor - "Yes, I need to compare the first half of the first string with the reverse order of the second half of the second string. Repeat this process to the first half of the second string and the second half of the first string."See full answer

Mridul J. - "We have a list of documents. We want to build an index that maps keywords to documents containing them. Then, given a query keyword, we can efficiently retrieve all matching documents. docs = [ "Python is great for data science", "C++ is a powerful language", "Python supports OOP and functional programming", "Weather today is sunny", "Weather forecast shows rain" ]"See full answer

Mahima M. - "class Solution: def lengthOfLIS(self, nums: List[int]) -> int: temp = [nums[0]] for num in nums: if temp[-1]< num: temp.append(num) else: index = bisect_left(temp,num) temp[index] = num return len(temp) "See full answer

Samuel M. - "function main(){ const v1=[2,3, 4, 10] const v2= [3,4 ,5,20, 23] return merge(v1,v2); } function merge(left, right){ const result=[]; while(left.length>0&& right.length>0){ if(left[0]0){ result=result.concat(left) } if(right.length>0){ result=result.concat(right) } return result; }"See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Michael B. - "Use an index, two pointers, and a set to keep track of elements that you've seen. pseudo code follows: for i, elem in enumerate(array): if elem in set return False if i > N: set.remove(array[i-N])"See full answer

Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

Anonymous Roadrunner - "from typing import List def traprainwater(height: List[int]) -> int: if not height: return 0 l, r = 0, len(height) - 1 leftMax, rightMax = height[l], height[r] res = 0 while l < r: if leftMax < rightMax: l += 1 leftMax = max(leftMax, height[l]) res += leftMax - height[l] else: r -= 1 rightMax = max(rightMax, height[r]) "See full answer