User Purchase Platform Problem
Description
LeetCode Problem 1127.
Table: Spending
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+-------------+---------+
| Column Name | Type |
+-------------+---------+
| user_id | int |
| spend_date | date |
| platform | enum |
| amount | int |
+-------------+---------+
The table logs the spendings history of users that make purchases from an online shopping website which has a desktop and a mobile application.
(user_id, spend_date, platform) is the primary key of this table.
The platform column is an ENUM type of ('desktop', 'mobile').
Write an SQL query to find the total number of users and the total amount spent using mobile only, desktop only and both mobile and desktop together for each date.
The query result format is in the following example:
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Spending table:
+---------+------------+----------+--------+
| user_id | spend_date | platform | amount |
+---------+------------+----------+--------+
| 1 | 2019-07-01 | mobile | 100 |
| 1 | 2019-07-01 | desktop | 100 |
| 2 | 2019-07-01 | mobile | 100 |
| 2 | 2019-07-02 | mobile | 100 |
| 3 | 2019-07-01 | desktop | 100 |
| 3 | 2019-07-02 | desktop | 100 |
+---------+------------+----------+--------+
Result table:
+------------+----------+--------------+-------------+
| spend_date | platform | total_amount | total_users |
+------------+----------+--------------+-------------+
| 2019-07-01 | desktop | 100 | 1 |
| 2019-07-01 | mobile | 100 | 1 |
| 2019-07-01 | both | 200 | 1 |
| 2019-07-02 | desktop | 100 | 1 |
| 2019-07-02 | mobile | 100 | 1 |
| 2019-07-02 | both | 0 | 0 |
+------------+----------+--------------+-------------+
On 2019-07-01, user 1 purchased using both desktop and mobile, user 2 purchased using mobile only and user 3 purchased using desktop only.
On 2019-07-02, user 2 purchased using mobile only, user 3 purchased using desktop only and no one purchased using both platforms.
MySQL Solution
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select dt.spend_date, dt.platform, ifnull(sum(amount), 0) as total_amount, count(user_id) as total_users
from
(select user_id, spend_date, if(ct = 2, "both", platform) as platform, amount
from
(select user_id, spend_date, count(user_id) as ct, platform, sum(amount) as amount
from spending
group by user_id, spend_date
) a
) b
right join
(select distinct(spend_date), 'desktop' platform from Spending
union
select distinct(spend_date), 'mobile' platform from Spending
union
select distinct(spend_date), 'both' platform from Spending
) dt
on b.spend_date = dt.spend_date and b.platform = dt.platform
group by spend_date, platform
order by spend_date
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