0550-game-play-analysis-iv.sql

-- 550. Game Play Analysis IV
-- Medium
-- https://leetcode.com/problems/game-play-analysis-iv

/*
Table: Activity
+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on someday using some device.

Write an SQL query to report the fraction of players that logged in again on the day after the day they first logged in, rounded to 2 decimal places. In other words, you need to count the number of players that logged in for at least two consecutive days starting from their first login date, then divide that number by the total number of players.
The query result format is in the following example.

Example 1:
Input: 
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-03-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+
Output: 
+-----------+
| fraction  |
+-----------+
| 0.33      |
+-----------+
Explanation: 
Only the player with id 1 logged back in after the first day he had logged in so the answer is 1/3 = 0.33
*/

SELECT 
  ROUND(
    COUNT(DISTINCT player_id) / (
      SELECT 
        COUNT(DISTINCT player_id) 
      FROM 
        Activity
    ), 
    2
  ) AS fraction 
FROM 
  Activity 
WHERE 
  (
    player_id, 
    DATE_SUB(event_date, INTERVAL 1 DAY)
  ) IN (
    SELECT 
      player_id, 
      MIN(event_date) AS first_login 
    FROM 
      Activity 
    GROUP BY 
      player_id
  );