DB 550. Game Play Analysis IV

Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)
Truncate table Activity
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-02', '6')
insert into Activity (player_id, device_id, event_date, games_played) values ('2', '3', '2017-06-25', '1')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')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


answer:
SELECT ROUND(COUNT(t2.player_id)/COUNT(t1.player_id),2) AS fraction
FROM
(SELECT player_id, MIN(event_date) AS first_login FROM Activity GROUP BY player_id) t1 LEFT JOIN Activity t2
ON t1.player_id = t2.player_id AND t1.first_login = t2.event_date - 1

ANSWER2:
# Write your MySQL query statement below
SELECT 
    ROUND(SUM(b.player_id is not null) / count(distinct a.player_id), 2) as fraction
FROM Activity a
LEFT JOIN
(
    SELECT player_id, date_add(min(event_date), interval 1 day) as event_date
    FROM Activity
    GROUP BY 1
) b
USING(player_id, event_date)

ANSWER3:
# Write your MySQL query statement below
SELECT ROUND(COUNT(t2.player_id)/COUNT(t1.player_id),2) AS fraction
FROM
(SELECT player_id, MIN(event_date) AS first_login FROM Activity GROUP BY player_id) t1 
LEFT JOIN Activity t2
ON t1.player_id = t2.player_id 
AND t1.first_login = t2.event_date - 1

# https://leetcode.com/problems/game-play-analysis-iv/solutions/315626/very-simple-mysql-solution/