Data Engineer Interview Questions

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

Lati K. - "I generally struggle with stakeholders and partners who doesn't communicate enough. Now it could be either they don't invest sufficient time and energy in doing so or at times they lack the skill sets to do so. In both the cases, the entire responsibility fell on the other person to dig deep into why someone is doing the way they are doing, reading into patterns and behaviour of their personality and adapting to those communication styles"See full answer

Temesgen B. - "Very Good Explanation Thanks For This Rely Good Explanation"See full answer

Anonymous Prawn - "Would be better to adjust resolution in the video player directly."See full answer

Bradley E. - "Here's a simpler solution: select u.username , count(p.postid) as countposts from posts as p join users as u on p.userid = u.userid where p.likes >= 100 group by 1 order by 2 desc, 1 asc limit 3 `"See full answer

Vysali K. - "Limit and rank() only works if there are no 2 employees with same salary ( which is okay for this use case) For the query to pass all the test results, we need to use dense_rank with ranked_employees as ( select id, firstname, lastname, salary, denserank() over(order by salary desc) as salaryrank from employees ) select id, firstname, lastname, salary from ranked_employees where salary_rank <= 3 `"See full answer

Sreeram reddy B. - "select employeename, employeeid, salary, department, DR from ( select employeename, employeeid, salary, dense_rank() over (partition by department order by salary desc) DR, department from employee ) where DR <=3 order by department, DR"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

Simran S. - "I recently led the development and implementation of a data analytics platform tailored for credit unions and mortgage companies, which was suffering from fragmented systems, inconsistent data fields across LOS platforms, and outdated reporting practices. Here's how I managed the full lifecycle: ✅ Initiation / Discovery Conducted executive interviews across five financial institutions to understand reporting and visibility gaps. Shadowed loan officers and underwriters"See full answer

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Anonymous Possum - "#inplace reversal without inbuilt functions def reverseString(s): chars = list(s) l, r = 0, len(s)-1 while l < r: chars[l],chars[r] = chars[r],chars[l] l += 1 r -= 1 reversed = "".join(chars) return reversed "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

Rishabh L. - "with empbysalary as ( select id, firstname, lastname, salary, department_id, rank() over (partition by department_id order by salary desc) as rnk from employees ) select d.name as department_name, e.id as employee_id, e.firstname, e.lastname, e.salary from empbysalary e join departments d on e.department_id=d.id where e.rnk=1 order by 1; `"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

Akshay D. - "SELECT u.user_id, u.user_name, u.email, ROUND(AVG(CASE WHEN b.status = 'Unmatched' THEN 1.0 ELSE 0 END), 2) AS avgunmatchedbookings FROM users u LEFT JOIN bookings b ON u.userid = b.userid GROUP BY u.user_id, u.user_name, u.email; `"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

Victor N. - "--country names are UPPERCASE but the table in the in the question showing lowercase. That's why it took me a while to figure it out until I ran the country column WITH RECURSIVE Hierarchy AS ( SELECT e.Emp_ID, CONCAT(e.FirstName, ' ', e.MiddleName, ' ', e.LastName) AS FullName, e.Manager_ID, 0 AS Level, CASE WHEN e.Country = 'IRELAND' THEN s.Salary * 1.09 WHEN e.Country = 'INDIA' THEN s.Salary * 0.012 ELSE s.Salary "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

Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer