Data Engineer Interview Questions

Berk C. - "High Level Architect Client v API Gateway v Object Storage v Message Queue v Worker v Database Client should can document with a web site or directly with API services. API Gateway should be used for upload document,get document info and state. Object storage should be used for original document and send event to Message Queue for starting. Message Queue is neccessary because there are millions of document should be process each time. Worker can get text from document with OCR. Database shoul"See full answer

Tiffany A. - "SELECT employees.first_name, managers.salary AS manager_salary FROM employees LEFT JOIN employees AS managers ON employees.manager_id = managers.id WHERE employees.salary > managers.salary `"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

Salome L. - "SELECT MIN(id) AS id, TRIM(LOWER(email)) AS cleaned_email FROM users GROUP BY cleaned_email ORDER BY id `"See full answer

Sreeram reddy B. - "user table - with userid, username, email, phonenumber, accountcreateddate exercises table - types of exercises - indoor walk, outdoor walk, running, stairs, cycling, swimming etc - exerciseid, exercisetype date table - date, day, month, year - with dateid Session table - userid, sessiondateid(linked to dateid in date table), exerciseid, distance covered, calories spent, starttime, endtime "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

Raj V. - "CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·"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

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

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

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

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

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

Hayatu H. - "What do all data scientists need to know about how to work with very large datasets? 37 Follow Request Answer More All related (39) Recommended 📷 Corrin Lakeland · Follow , M.S. Data Science, University of St. Thomas, St. Paul (2018)6yData Science consultant and managerUpvoted by[Tom Halloin](https://www.quora"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

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

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

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