Data Engineer Interview Questions

David I. - "WITH filtered_posts AS ( SELECT p.user_id, p.issuccessfulpost FROM post p WHERE p.postdate >= '2023-11-01' AND p.postdate < '2023-12-01' ), post_summary AS ( SELECT pu.user_type, COUNT(*) AS post_attempt, SUM(CASE WHEN fp.issuccessfulpost = 1 THEN 1 ELSE 0 END) AS post_success FROM filtered_posts fp JOIN postuser pu ON fp.userid = pu.user_id GROUP BY pu.user_type ) SELECT user_type, post_success, post_attempt, CAST(postsuccess AS FLOAT) / postattempt AS postsuccessrate FROM po"See full answer

Zakery K. - "Conflict is a GREAT opportunity to really demonstrate that you care about someone and, through effective conflict resolution, build stronger authentic relationships with the people you work with. When faced with conflict, I prioritize understanding all perspectives involved. I start by actively listening to the other parties: asking clarifying questions to pinpoint the source of the conflict, reflecting back what I'm hearing to make sure I understand them correctly, and ultimately identify"See full answer

Amney M. - "it is really good explanation thanks it is really good explanation thanks"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

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and mortgage companies, which was suffering from fragmented systems, inconsistent data fields across LOS platforms, and outdated reporting practices.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

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

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

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

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

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

Anisha S. - "SELECT d.name as departmentname,e.id as employeeid,e.firstname,e.lastname,MAX(e.salary) as salary FROM employees e LEFT JOIN departments d ON e.department_id=d.id GROUP BY department_name ORDER BY department_name;"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

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

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

Hayatu H. - "How do you find consecutive days for login (MySQL, SQL, date, subquery, MySQL 5.7, development)? 1 Follow Request Answer More All related (34) Recommended 📷 Trausti Thor Johannsson · Follow Been using MySQL for more than 16 yearsDec 27 There are functions like DATEDIFF but there are also BETWE"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