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Introduction To SQL
SQL stands for Structured Query Language. SQL is used to create, remove, alter the database and database objects in a database management system and to store, retrieve, update the data in a database. SQL is a standard language for creating, accessing, manipulating database management system. SQL works for all modern relational database management systems, like SQL Server, Oracle, MySQL, etc.
- It is a standard language for Relational Database System. It enables a user to create, read, update and delete relational databases and tables.
- All the RDBMS like MySQL, Informix, Oracle, MS Access and SQL Server use SQL as their standard database language.
- SQL allows users to query the database in a number of ways, using English-like statements.
Link for the Problem – 15 Days of Learning SQL – Hacker Rank Solution
15 Days of Learning SQL – Hacker Rank Solution
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
15 Days of Learning SQL – Hacker Rank 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
;
