LeetCode--1934. 确认率

1 题目描述

表: Signups

+----------------+----------+
| Column Name    | Type     |
+----------------+----------+
| user_id        | int      |
| time_stamp     | datetime |
+----------------+----------+

User_id是该表的主键
每一行都包含ID为user_id的用户的注册时间信息

表: Confirmations

+----------------+----------+
| Column Name    | Type     |
+----------------+----------+
| user_id        | int      |
| time_stamp     | datetime |
| action         | ENUM     |
+----------------+----------+

(user_id, time_stamp)是该表的主键
user_id是一个引用到注册表的外键
action是类型为(‘confirmed’, ‘timeout’)的ENUM
该表的每一行都表示ID为user_id的用户在time_stamp请求了一条确认消息，该确认消息要么被确认(‘confirmed’), 要么被过期(‘timeout’)
用户的 确认率 是 ‘confirmed’ 消息的数量除以请求的确认消息的总数。没有请求任何确认消息的用户的确认率为 0 。确认率四舍五入到 小数点后两位
编写一个SQL查询来查找每个用户的 确认率
以 任意顺序 返回结果表

2 测试用例

输入：
Signups 表:

+---------+---------------------+
| user_id | time_stamp          |
+---------+---------------------+
| 3       | 2020-03-21 10:16:13 |
| 7       | 2020-01-04 13:57:59 |
| 2       | 2020-07-29 23:09:44 |
| 6       | 2020-12-09 10:39:37 |
+---------+---------------------+

Confirmations 表:

+---------+---------------------+-----------+
| user_id | time_stamp          | action    |
+---------+---------------------+-----------+
| 3       | 2021-01-06 03:30:46 | timeout   |
| 3       | 2021-07-14 14:00:00 | timeout   |
| 7       | 2021-06-12 11:57:29 | confirmed |
| 7       | 2021-06-13 12:58:28 | confirmed |
| 7       | 2021-06-14 13:59:27 | confirmed |
| 2       | 2021-01-22 00:00:00 | confirmed |
| 2       | 2021-02-28 23:59:59 | timeout   |
+---------+---------------------+-----------+

输出:

+---------+-------------------+
| user_id | confirmation_rate |
+---------+-------------------+
| 6       | 0.00              |
| 3       | 0.00              |
| 7       | 1.00              |
| 2       | 0.50              |
+---------+-------------------+

解释:

• 用户 6 没有请求任何确认消息, 确认率为 0
• 用户 3 进行了 2 次请求, 都超时了, 确认率为 0
• 用户 7 提出了 3 个请求, 所有请求都得到了确认, 确认率为 1
• 用户 2 做了 2 个请求, 其中一个被确认, 另一个超时, 确认率为 1 / 2 = 0.5

3 解题思路

1. 需要考虑在Signups 中存在的用户,但Confirmations 中没有记录, 所以使用 Signups左外连接Confirmations 表, 并对user_id分组统计用户在Confirmations 中的请求数据条数(user_confirmed_count)和已确认数据条数(user_count)

select s.user_id, c.action,
       sum(if(c.action = 'confirmed', 1, 0)) over (partition by c.user_id) as user_confirmed_count,
       count(*) over (partition by c.user_id)                              as user_count
from Signups as s
         left join Confirmations as c on s.user_id = c.user_id;

查询结果

+-------+---------+--------------------+----------+
|user_id|action   |user_confirmed_count||
+-------+---------+--------------------+----------+
|7      |confirmed|3                   |3         |
|7      |confirmed|3                   |3         |
|7      |confirmed|3                   |3         |
|2      |timeout  |1                   |2         |
|2      |confirmed|1                   |2         |
|3      |timeout  |0                   |2         |
|3      |timeout  |0                   |2         |
|6      |null     |0                   |1         |
+-------+---------+--------------------+----------+

2. 将每个用户的确认数据条数(user_confirmed_count) / 总条数(user_count), 就能能到确认率, 再对确认率进行四舍五入,保留两位小数, 对user_id进行去重

select distinct s.user_id,
                round((sum(if(c.action = 'confirmed', 1, 0)) over (partition by c.user_id)) /
                      (count(*) over (partition by c.user_id)), 2) as confirmation_rate
from Signups as s
         left join Confirmations as c on s.user_id = c.user_id;

查询结果

+-------+-----------------+
|user_id|confirmation_rate|
+-------+-----------------+
|6      |0.00             |
|2      |0.50             |
|3      |0.00             |
|7      |1.00             |
+-------+-----------------+