SQL-50-Leetcode-Problems/1633. Percentage of Users Attended a Contest.sql at main · evanmathew/SQL-50-Leetcode-Problems · GitHub

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-- Problem:

/*
Table: Users

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| user_id     | int     |
| user_name   | varchar |
+-------------+---------+
user_id is the primary key (column with unique values) for this table.
Each row of this table contains the name and the id of a user.


Table: Register

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| contest_id  | int     |
| user_id     | int     |
+-------------+---------+
(contest_id, user_id) is the primary key (combination of columns with unique values) for this table.
Each row of this table contains the id of a user and the contest they registered into.


Write a solution to find the percentage of the users registered in each contest rounded to two decimals.

Return the result table ordered by percentage in descending order. In case of a tie, order it by contest_id in ascending order.

The result format is in the following example.


Example 1:

Input:
Users table:
+---------+-----------+
| user_id | user_name |
+---------+-----------+
| 6       | Alice     |
| 2       | Bob       |
| 7       | Alex      |
+---------+-----------+
Register table:
+------------+---------+
| contest_id | user_id |
+------------+---------+
| 215        | 6       |
| 209        | 2       |
| 208        | 2       |
| 210        | 6       |
| 208        | 6       |
| 209        | 7       |
| 209        | 6       |
| 215        | 7       |
| 208        | 7       |
| 210        | 2       |
| 207        | 2       |
| 210        | 7       |
+------------+---------+
Output:
+------------+------------+
| contest_id | percentage |
+------------+------------+
| 208        | 100.0      |
| 209        | 100.0      |
| 210        | 100.0      |
| 215        | 66.67      |
| 207        | 33.33      |
+------------+------------+
Explanation:
All the users registered in contests 208, 209, and 210. The percentage is 100% and we sort them in the answer table by contest_id in ascending order.
Alice and Alex registered in contest 215 and the percentage is ((2/3) * 100) = 66.67%
Bob registered in contest 207 and the percentage is ((1/3) * 100) = 33.33%
*/

-------------------------------------------------------------------------------

-- Solution:

SELECT r.contest_id, ROUND((COUNT(DISTINCT r.user_id) / (SELECT COUNT(DISTINCT user_id) FROM Users)) * 100, 2) AS percentage
FROM Users u
GROUP BY r.contest_id
ORDER BY percentage DESC, contest_id ASC