LeetCode 코딩 테스트 - Game Play Analysis IV(LV.Medium)

 

 

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 (combination of columns with unique values) 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 a solution 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 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

 

# 쿼리를 작성하는 목표, 확인할 지표 : 전체 중에서 첫날과 둘째날 모두 접속 기록이 있는 사람 비율 구하기 / event_date, player_id
# 쿼리를 계산하는 방법 : 1. 전체 player 수 구하기 -> 2. 각 player 별 첫날 구하기 -> 3. 첫날과 둘째날 모두 접속한 사람의 수 구하기 -> 4. 비율 계산
# 데이터의 기간 : x
# 사용할 테이블 : activity
# JOIN KEY : x
# 데이터 특징 : x
WITH total_cnt AS (
    # 1
    SELECT
        COUNT(DISTINCT player_id) AS cnt
    FROM activity
), first_day AS (
    # 2
    SELECT
        player_id,
        MIN(event_date) AS first_day
    FROM activity
    GROUP BY
        player_id
), consecutive_cnt AS (
    # 3
    SELECT
        COUNT(*) AS consecutive_cnt
    FROM activity AS a
    INNER JOIN first_day AS f
    ON a.player_id = f.player_id
    WHERE
        f.first_day + INTERVAL 1 DAY = a.event_date
)
# 4
SELECT
    ROUND(consecutive_cnt / (SELECT cnt FROM total_cnt), 2) AS fraction
FROM consecutive_cnt