Skip to main content

Data Analyst Coding Interview Questions

Review this list of 26 Coding Data Analyst interview questions and answers verified by hiring managers and candidates.
  • 76 answers
    Video answer for 'Employee Earnings.'
    +69

    "select e.firstname as firstname, m.salary as manager_salary from employees e join employees m on e.manager_id = m.id where e.salary > m.salary; `"

    Ravi K. - "select e.firstname as firstname, m.salary as manager_salary from employees e join employees m on e.manager_id = m.id where e.salary > m.salary; `"See full answer

    Data Analyst
    Coding
    +4 more
  • Meta logoAsked at Meta 
    13 answers
    +8

    "Clarifying Points Number of users: unique user id “More than three people” we have user id for the call recipient, and we want number of unique recipients called. We would not count each instance of calling someone. Last week: last 7 days from today SELECT User_id, COUNT(distinct calleduserid) as userscalledcount FROM calls WHERE calldate between currentdate - interval ‘7 day’ and current_date Group by user_id HAVING COUNT(distinct calleduserid) >= 3 This return"

    E B. - "Clarifying Points Number of users: unique user id “More than three people” we have user id for the call recipient, and we want number of unique recipients called. We would not count each instance of calling someone. Last week: last 7 days from today SELECT User_id, COUNT(distinct calleduserid) as userscalledcount FROM calls WHERE calldate between currentdate - interval ‘7 day’ and current_date Group by user_id HAVING COUNT(distinct calleduserid) >= 3 This return"See full answer

    Data Analyst
    Coding
    +3 more
  • LinkedIn logoAsked at LinkedIn 
    36 answers
    +31

    "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"

    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

    Data Analyst
    Coding
    +4 more
  • 47 answers
    +43

    "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 `"

    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

    Data Analyst
    Coding
    +3 more
  • 4 answers
    Video answer for 'SQL Stored Procedures'
    +1

    "it is really good explanation thanks it is really good explanation thanks"

    Amney M. - "it is really good explanation thanks it is really good explanation thanks"See full answer

    Data Analyst
    Coding
    +4 more
  • 🧠 Want an expert answer to a question? Saving questions lets us know what content to make next.

  • 69 answers
    +63

    "SELECT id, first_name, last_name, salary FROM employees ORDER BY salary DESC LIMIT 3 `"

    Kavi S. - "SELECT id, first_name, last_name, salary FROM employees ORDER BY salary DESC LIMIT 3 `"See full answer

    Data Analyst
    Coding
    +3 more
  • Amazon logoAsked at Amazon 
    57 answers
    Video answer for 'Merge Intervals'
    +49

    "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"

    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

    Data Analyst
    Coding
    +6 more
  • Tesla logoAsked at Tesla 
    35 answers
    +32

    "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; `"

    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

    Data Analyst
    Coding
    +4 more
  • 25 answers
    +22

    "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 `"

    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

    Data Analyst
    Coding
    +3 more
  • 28 answers
    +21

    "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; `"

    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

    Data Analyst
    Coding
    +3 more
  • 22 answers
    +17

    "--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 "

    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

    Data Analyst
    Coding
    +3 more
  • 28 answers
    +25

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

    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

    Data Analyst
    Coding
    +3 more
  • 14 answers
    +11

    "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"

    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

    Data Analyst
    Coding
    +3 more
  • Google logoAsked at Google 
    5 answers
    +2

    "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; `"

    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

    Data Analyst
    Coding
    +4 more
  • Google logoAsked at Google 
    27 answers
    +24

    "with cte as (select ts.employee_id, e.name, t.id as test_id, max(distinct ts.score) as total from test_results as ts join tests as t on ts.test_id = t.id join employees as e on ts.employee_id = e.id group by ts.employee_id, e.name, t.id) select employee_id, name as employee_name, sum(total) as total_score from cte group by employee_id, employee_name order by total_score desc, employee_id asc ;"

    Christian B. - "with cte as (select ts.employee_id, e.name, t.id as test_id, max(distinct ts.score) as total from test_results as ts join tests as t on ts.test_id = t.id join employees as e on ts.employee_id = e.id group by ts.employee_id, e.name, t.id) select employee_id, name as employee_name, sum(total) as total_score from cte group by employee_id, employee_name order by total_score desc, employee_id asc ;"See full answer

    Data Analyst
    Coding
    +3 more
  • Amazon logoAsked at Amazon 
    6 answers

    "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"

    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

    Data Analyst
    Coding
    +4 more
  • 14 answers
    +10

    " select user_id, b.marketing_channel from user_sessions a Left join attribution b on b.sessionid = a.sessionid group by 1,2 HAVING sum(purchasevalue)>100 and min(adclick_timestamp) `"

    G B. - " select user_id, b.marketing_channel from user_sessions a Left join attribution b on b.sessionid = a.sessionid group by 1,2 HAVING sum(purchasevalue)>100 and min(adclick_timestamp) `"See full answer

    Data Analyst
    Coding
    +3 more
  • 15 answers
    +12

    " with youngsuccrate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as yascrate from post where userid in (select userid from post_user where age between 0 and 18) group by post_month ), nonyoungsucc_rate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as nonyasc_rate from post where user_id in (select"

    Bhavna S. - " with youngsuccrate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as yascrate from post where userid in (select userid from post_user where age between 0 and 18) group by post_month ), nonyoungsucc_rate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as nonyasc_rate from post where user_id in (select"See full answer

    Data Analyst
    Coding
    +3 more
  • Add answer
    Video answer for 'Analyze Monthly Customer Transactions'
    Data Analyst
    Coding
    +3 more
  • 13 answers
    +10

    "SELECT upsellcampaignid, COUNT(DISTINCT trans.userid) AS eligibleusers FROM campaign JOIN "transaction" AS trans ON transactiondate BETWEEN datestart AND date_end JOIN user ON trans.userid = user.userid WHERE iseligibleforupsellcampaign = 1 GROUP BY upsellcampaignid `"

    Alina G. - "SELECT upsellcampaignid, COUNT(DISTINCT trans.userid) AS eligibleusers FROM campaign JOIN "transaction" AS trans ON transactiondate BETWEEN datestart AND date_end JOIN user ON trans.userid = user.userid WHERE iseligibleforupsellcampaign = 1 GROUP BY upsellcampaignid `"See full answer

    Data Analyst
    Coding
    +3 more
Showing 1-20 of 26
Exponent

Get updates in your inbox with the latest tips, job listings, and more.

Follow Us

Products
Courses
Interview Questions
Interview Experiences
Popular articles
Guides
Coaching
For Partners
Company
Exponent © 2026
Terms of Service | Privacy