15 Days of Learning SQL HackerRank SQL Solution in MYSQL

Problem:-

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

On March 01, 2016 hackers , , , and  made submissions. There are  unique hackers who made at least one submission each day. As each hacker made one submission,  is considered to be the hacker who made maximum number of submissions on this day. The name of the hacker is Angela.

On March 02, 2016 hackers , , and  made submissions. Now  and  were the only ones to submit every day, so there are  unique hackers who made at least one submission each day.  made  submissions, and name of the hacker is Michael.

On March 03, 2016 hackers , , and  made submissions. Now  and  were the only ones, so there are  unique hackers who made at least one submission each day. As each hacker made one submission so  is considered to be the hacker who made maximum number of submissions on this day. The name of the hacker is Angela.

On March 04, 2016 hackers , , , and  made submissions. Now  and  only submitted each day, so there are  unique hackers who made at least one submission each day. As each hacker made one submission so  is considered to be the hacker who made maximum number of submissions on this day. The name of the hacker is Angela.

On March 05, 2016 hackers , ,  and  made submissions. Now  only submitted each day, so there is only  unique hacker who made at least one submission each day.  made  submissions and name of the hacker is Frank.

On March 06, 2016 only  made submission, so there is only  unique hacker who made at least one submission each day.  made  submission and name of the hacker is Angela.

Solution:-

SELECT t1.submission_date, hkr_cnt, t2.hacker_id, name
FROM (SELECT p1.submission_date, 
             COUNT(DISTINCT p1.hacker_id) AS hkr_cnt
      FROM (SELECT submission_date, hacker_id, 
                   @h_rnk := CASE WHEN @h_grp != hacker_id THEN 1 ELSE @h_rnk+1 END AS hacker_rank,
                   @h_grp := hacker_id AS hacker_group
            FROM (SELECT DISTINCT submission_date, hacker_id 
                  FROM submissions
                  ORDER BY hacker_id, submission_date) AS a, 
                 (SELECT @h_rnk := 1, @h_grp := 0) AS r) AS p1
      JOIN (SELECT submission_date, 
                   @d_rnk := @d_rnk + 1 AS date_rank
            FROM (SELECT DISTINCT submission_date
                  FROM submissions 
                  ORDER BY submission_date) AS b, 
                 (SELECT @d_rnk := 0) r) AS p2
      ON p1.submission_date = p2.submission_date 
         AND hacker_rank = date_rank
      GROUP BY p1.submission_Date) AS t1
JOIN (SELECT submission_date, hacker_id, sub_cnt,
             @s_rnk := CASE WHEN @d_grp != submission_date THEN 1 ELSE @s_rnk+1 END AS max_rnk,
             @d_grp := submission_date AS date_group
      FROM (SELECT submission_date, hacker_id, COUNT(*) AS sub_cnt
            FROM submissions AS s
            GROUP BY submission_date, hacker_id
            ORDER BY submission_date, sub_cnt DESC, hacker_id) AS c,
           (SELECT @s_rnk := 1, @d_grp := 0) AS r) AS t2                            
ON t1.submission_date = t2.submission_date AND max_rnk = 1
JOIN hackers AS h ON h.hacker_id = t2.hacker_id            
ORDER BY t1.submission_date

;