Meta Data Engineer Interview Questions

Sreenisha S. - "Let me tell you about a time where a website I managed suddenly showed slow performance and the mistake on our side was it was unnoticed until a user reported the issue to management. As a PM for that project, I took full responsibility of the situation and worked with the engineering team to quickly resolve it. This mistake taught me the importance of focusing and monitoring non functional requirements as well in addition to new feature development /adoption where I was mostly spending my time"See full answer

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

Bipin R. - "As you know, this is the most important question for any interview. Here is a structure I like to follow, Start with 'I'm currently a SDE/PM/TPM etc with XYZ company.... ' Mention how you got into PM/TPM/SDE field (explaining your journey) Mention 1 or 2 accomplishments Mention what you do outside work (blogging, volunteer etc) Share why are you looking for a new role Ask the interviewer if they have any questions or will like to dive deep into any of your experience"See full answer

KAI - "Answer: select fromcaller, count(DISTINCT tocallee) as num_calls from calls group by fromcaller having count(DISTINCT tocallee) >= 3 Setup: CREATE TABLE calls ( from_caller VARCHAR(20), to_callee VARCHAR(20) ); INSERT INTO calls (fromcaller, tocallee) VALUES ('Alice', 'Bob'), ('Charlie', 'Dave'), ('Alice', 'Frank'), ('Charlie', 'Heidi'), ('Charlie', 'Judy'); "See full answer

Anonymous Roadrunner - "def is_palindrome(s: str) -> bool: new = '' for a in s: if a.isalpha() or a.isdigit(): new += a.lower() return (new == new[::-1]) debug your code below print(is_palindrome('abcba')) `"See full answer

Malay K. - "For any project based questions, it is important to structure your response clearly, showcasing your thought process, technical skills, problem-solving abilities, and how your work added value. Besides the STAR method, you can also use this kind of framework: 1. Start by selecting a relevant project (related to the role) Give the project background and what specific problem it solved. 2. Align the project's objective and your role Be specific about your role: were you the le"See full answer

Abhinav M. - "Always assume good intentions on the part of both parties when resolving conflicts. Then proceed with a STAR example."See full answer

Bala G. - "SELECT s.Sale_Date, SUM(si.Quantity * si.SalePrice) AS TotalRevenue FROM Sales s JOIN SaleItems si ON s.SaleID = si.Sale_ID GROUP BY s.Sale_Date ORDER BY s.Sale_Date; "See full answer

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

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

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

Scott S. - " A couple of years ago, we were working on a project to integrate a new third-party data feed into our existing data processing pipeline. This data feed was critical for enhancing our trading algorithms with more comprehensive market data. Given the tight timeline and high stakes, I decided to push for a rapid implementation. In my eagerness to meet the deadline, I underestimated the complexity of integrating this new data feed. I did not allocate sufficient time for thorough testing and valida"See full answer

Anonymous Roadrunner - "from typing import List def two_sum(nums: List[int], target: int) -> List[int]: prevMap = {} for i, n in enumerate(nums): diff = target - n if diff in prevMap: return [prevMap[diff], i] else: prevMap[n] = i return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"See full answer

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

Ankita G. - "Merge Sort"See full answer

Ramachandra N. - "def mergeTwoListsRecursive(l1, l2): if not l1 or not l2: return l1 or l2 if l1.val < l2.val: l1.next = mergeTwoListsRecursive(l1.next, l2) return l1 else: l2.next = mergeTwoListsRecursive(l1, l2.next) return l2 "See full answer

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"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

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer