# Game Play Analysis V Problem

## Description

LeetCode Problem 1097.

Table: Activity

``````1
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+--------------+---------+
| 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 game.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.
``````

We define the install date of a player to be the first login day of that player.

We also define day 1 retention of some date X to be the number of players whose install date is X and they logged back in on the day right after X, divided by the number of players whose install date is X, rounded to 2 decimal places.

Write an SQL query that reports for each install date, the number of players that installed the game on that day and the day 1 retention.

The query result format is in the following example:

``````1
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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         | 2016-03-01 | 0            |
| 3         | 4         | 2016-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result table:
+------------+----------+----------------+
| install_dt | installs | Day1_retention |
+------------+----------+----------------+
| 2016-03-01 | 2        | 0.50           |
| 2017-06-25 | 1        | 0.00           |
+------------+----------+----------------+
Player 1 and 3 installed the game on 2016-03-01 but only player 1 logged back in on 2016-03-02 so the day 1 retention of 2016-03-01 is 1 / 2 = 0.50
Player 2 installed the game on 2017-06-25 but didn't log back in on 2017-06-26 so the day 1 retention of 2017-06-25 is 0 / 1 = 0.00
``````

## MySQL Solution

``````1
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select t1.install_date as install_dt, count(t1.install_date) as installs,
round(count(t2.event_date) / count(*), 2) as Day1_retention
from (
select player_id, min(event_date) as install_date
from Activity
group by 1
) t1
left join Activity t2
on date_add(t1.install_date, interval 1 day) = t2.event_date
and t1.player_id = t2.player_id
group by 1
order by 1
``````