15 Days of Learning SQL — HackerRank Advanced SQL Joins

Julia conducted a days of learning SQL contest. The start date of the contest was March 01, 2016 and the end date was March 15, 2016.

Write a query to print total number of unique hackers who made at least submission each day (starting on the first day of the contest), and find the hacker_id and name of the hacker who made maximum number of submissions each day. If more than one such hacker has a maximum number of submissions, print the lowest hacker_id. The query should print this information for each day of the contest, sorted by the date.

Input Format

The following tables hold contest data:

• Hackers: The hacker_id is the id of the hacker, and name is the name of the hacker.

• Submissions: The submission_date is the date of the submission, submission_id is the id of the submission, hacker_id is the id of the hacker who made the submission, and score is the score of the submission.

Sample Input

For the following sample input, assume that the end date of the contest was March 06, 2016.

Hackers Table:

Submissions Table:

Sample Output

2016-03-01 4 20703 Angela
2016-03-02 2 79722 Michael
2016-03-03 2 20703 Angela
2016-03-04 2 20703 Angela
2016-03-05 1 36396 Frank
2016-03-06 1 20703 Angela

Explanation

Solution

SELECT sd, 
       cnt, 
       h.hacker_id, 
       h.name 
FROM  (SELECT sd, 
              (SELECT Count(DISTINCT h.hacker_id) 
               FROM   hackers h 
                      JOIN submissions s 
                        ON s.hacker_id = h.hacker_id 
               WHERE  s.submission_date <= sd 
                      AND (SELECT Count(DISTINCT submission_date) 
                           FROM   hackers hh 
                                  JOIN submissions ss 
                                    ON ss.hacker_id = hh.hacker_id 
                           WHERE  hh.hacker_id = h.hacker_id 
                                  AND ss.submission_date <= sd) = 
                          1 + Datediff(sd, '2016-03-01')) AS cnt, 
              (SELECT h.hacker_id 
               FROM   hackers h 
                      JOIN submissions s 
                        ON s.hacker_id = h.hacker_id 
               WHERE  s.submission_date = sd 
               GROUP  BY h.hacker_id 
               ORDER  BY Count(s.submission_id) DESC, 
                         h.hacker_id 
               LIMIT  1)                                  AS hacker_id, 
              (SELECT Count(s.submission_id) 
               FROM   hackers h 
                      JOIN submissions s 
                        ON s.hacker_id = h.hacker_id 
               WHERE  s.submission_date = sd 
               GROUP  BY h.hacker_id 
               ORDER  BY Count(s.submission_id) DESC, 
                         h.hacker_id 
               LIMIT  1)                                  AS hacker_cnt 
       FROM   (SELECT submission_date sd 
               FROM   submissions 
               GROUP  BY submission_date) AS date_tbl) AS TBL2 
      JOIN hackers h 
        ON h.hacker_id = TBL2.hacker_id 
ORDER  BY sd

Expected output

2016-03-01 112 81314 Denise 
2016-03-02 59 39091 Ruby 
2016-03-03 51 18105 Roy 
2016-03-04 49 533 Patrick 
2016-03-05 49 7891 Stephanie 
2016-03-06 49 84307 Evelyn 
2016-03-07 35 80682 Deborah 
2016-03-08 35 10985 Timothy 
2016-03-09 35 31221 Susan 
2016-03-10 35 43192 Bobby 
2016-03-11 35 3178 Melissa 
2016-03-12 35 54967 Kenneth 
2016-03-13 35 30061 Julia 
2016-03-14 35 32353 Rose 
2016-03-15 35 27789 Helen