Hacker Rank SQL Challenge — 15 Days of Learning SQL

— — This is the challenge comes from Hacker Rank SQL test, please state the source when referring to this article

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.

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This problem is a little bit tricky as you should get the aggregated value

Here is the solution

SELECT SUBMISSION_DATE,
(SELECT COUNT(DISTINCT HACKER_ID)  
 FROM SUBMISSIONS S2  
 WHERE S2.SUBMISSION_DATE = S1.SUBMISSION_DATE AND    
(SELECT COUNT(DISTINCT S3.SUBMISSION_DATE) 
 FROM SUBMISSIONS S3 WHERE S3.HACKER_ID = S2.HACKER_ID AND S3.SUBMISSION_DATE < S1.SUBMISSION_DATE) = DATEDIFF(S1.SUBMISSION_DATE , '2016-03-01')),
(SELECT HACKER_ID FROM SUBMISSIONS S2 WHERE S2.SUBMISSION_DATE = S1.SUBMISSION_DATE 
GROUP BY HACKER_ID ORDER BY COUNT(SUBMISSION_ID) DESC, HACKER_ID LIMIT 1) AS TMP,
(SELECT NAME FROM HACKERS WHERE HACKER_ID = TMP)
FROM
(SELECT DISTINCT SUBMISSION_DATE FROM SUBMISSIONS) S1
GROUP BY SUBMISSION_DATE;

Here we could break down the problem into a few small problems,

1. the number of people who has made consecutive submissions in the past few days
2. among the people who had make consecutive submission , who submit the most amount of data

To solve the first part

(SELECT COUNT(DISTINCT HACKER_ID)  
 FROM SUBMISSIONS S2  
 WHERE S2.SUBMISSION_DATE = S1.SUBMISSION_DATE AND    
(SELECT COUNT(DISTINCT S3.SUBMISSION_DATE) 
 FROM SUBMISSIONS S3 WHERE S3.HACKER_ID = S2.HACKER_ID AND S3.SUBMISSION_DATE < S1.SUBMISSION_DATE) = DATEDIFF(S1.SUBMISSION_DATE , '2016-03-01'))

Here we select the number of distinct hacker whose on certain date equal than the number of days the contest start

And the second part

(SELECT HACKER_ID FROM SUBMISSIONS S2 WHERE S2.SUBMISSION_DATE = S1.SUBMISSION_DATE 
GROUP BY HACKER_ID ORDER BY COUNT(SUBMISSION_ID) DESC, HACKER_ID LIMIT 1) AS TMP,
(SELECT NAME FROM HACKERS WHERE HACKER_ID = TMP)FROM
(SELECT DISTINCT SUBMISSION_DATE FROM SUBMISSIONS) S1
GROUP BY SUBMISSION_DATE;

and incorporate these two selection in the main part of selection