Data Engineer Interview Questions

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

Jeremy D. - "Any cycle would cause the prerequisite to be greater than the course. This passes all the tests: function canFinish(_numCourses, prerequisites) { for (const [a, b] of prerequisites) { if (b > a) return false } return true } `"See full answer

Karthik R. - "This is yet another classic case of evolution of data landscape to account for diversities in the data formats sacrificing restrictive but key components at first and added later to make the solution more effective. Data warehouse -> Data Lake -> Data Lakehouse (Data Lake + Data Warehouse) Data warehouse - A solution to store data in central place (analytics (read) heavy) with stringent schema (structured). Very useful for historical queries and analytics. Schema on write check. Only used for"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

Abhishek K. - "-- Write your query here with cte as ( SELECT e.Emp_ID, e.First_Name, e.Middle_Name, e.Last_Name, e.Manager_ID, e.Country, s.Salary as original_salary, CASE WHEN Country = 'IRELAND' THEN s.Salary*1.09 WHEN Country = 'INDIA' THEN s.Salary*0.012 ELSE s.Salary END AS Salary FROM tbl_Employee e LEFT JOIN tbl_EmployeeSalary s ON e.Emp_ID = s.Emp"See full answer

Nidhi S. - "There are couple of reasons for it - Kind of role : Its a product manager role loaded with analytical work, So working with data in stringent regulatory guideline make it more exciting and thrilling. Location & industry is like - Cherry on the cake, Bangalore weather and BFI is at its all time peak as people spending behavior is changing continuously, it will be interesting to see big giants like visa are managing it."See full answer

Daniel C. - "select name, stock from products p left join transactions t on p.id = t.product_id order by date desc limit 1"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

Evan R. - "The user table no longer exists as expected - I get an error that user does not contain user_id. Note that querying the table results in only user:swuoevkivrjfta select * FROM user `"See full answer

Yashasvi V. - "WITH RECURSIVE fibonacci_series AS ( SELECT 1 AS n, 0 AS fib1, 1 AS fib2 UNION ALL SELECT n + 1 AS n, fib2 AS fib1, fib1 + fib2 AS fib2 FROM fibonacci_series WHERE n < 20 -- Limit the series to 20 numbers ) SELECT n, fib1 AS fib FROM fibonacci_series ORDER BY n; `"See full answer

Nilanjan D. - "Firstly, congratulations to both the interviewer and interviewee. This was a great learning experience However, being a Full Stack engineer and I was having the following suggestions around the Data Model - Driver & Approval can be two different tables Approval & Document - Approval can be a tuple of (userid,documentid) - comments against a rejection (marks the document which triggers rejection)In this way we can capture the entire history of approval workflow (initiate/pending/appr"See full answer

Victor H. - "There is a faster approach that solves the problem in O(n) time: def find_duplicates(arr1, arr2): arr1 = set(arr1) res = [] for num in arr2: if num in arr1: res.append(num) return res `"See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Mayank S. - "[2:53 pm, 02/12/2021] Mayank: Before we deep dive into brainstorming the solution I go ahead and make few assumptions for clarifying questions: Step-1- Framing a problem 🎯Why users go to gym? To relieve stress by doing exercise To maintain their body To reduce their weight To remain active To make good physique 🎯Who goes to gym? Couples Group of friends Individuals Here we are trying to design high tech gym so it means we are looking to create good experience-and"See full answer

Peter W. - "In the question it says: "above the overall average total posts", which to me implying a >, yet in the solution it uses >= Caused me 1 hr to find out. plz fix"See full answer

Erjan G. - "1) select avg(session) from table where session> 180 2) select round(sessiontime/300)*300 as sessionbin, count() as sessioncount from table group by round(sessiontime/300)300 order by session_bin 3) SELECT t1.country AS country_a, t2.country AS country_b FROM ( SELECT country, COUNT(*) AS session_count FROM yourtablename GROUP BY country ) AS t1 JOIN ( SELECT country, COUNT(*) AS session_count FROM yourtablename `GROUP BY countr"See full answer