Customer Order Frequency Problem

Description

LeetCode Problem 1511.

Table: Customers

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| country       | varchar |
+---------------+---------+
customer_id is the primary key for this table.
This table contains information of the customers in the company.
``````

Table: Product

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| description   | varchar |
| price         | int     |
+---------------+---------+
product_id is the primary key for this table.
This table contains information of the products in the company.
price is the product cost.
``````

Table: Orders

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| customer_id   | int     |
| product_id    | int     |
| order_date    | date    |
| quantity      | int     |
+---------------+---------+
order_id is the primary key for this table.
This table contains information on customer orders.
customer_id is the id of the customer who bought "quantity" products with id "product_id".
Order_date is the date in format ('YYYY-MM-DD') when the order was shipped.
``````

Write an SQL query to report the customer_id and customer_name of customers who have spent at least \$100 in each month of June and July 2020.

Return the result table in any order.

The query result format is in the following example.

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Customers
+--------------+-----------+-------------+
| customer_id  | name      | country     |
+--------------+-----------+-------------+
| 1            | Winston   | USA         |
| 2            | Jonathan  | Peru        |
| 3            | Moustafa  | Egypt       |
+--------------+-----------+-------------+

Product
+--------------+-------------+-------------+
| product_id   | description | price       |
+--------------+-------------+-------------+
| 10           | LC Phone    | 300         |
| 20           | LC T-Shirt  | 10          |
| 30           | LC Book     | 45          |
| 40           | LC Keychain | 2           |
+--------------+-------------+-------------+

Orders
+--------------+-------------+-------------+-------------+-----------+
| order_id     | customer_id | product_id  | order_date  | quantity  |
+--------------+-------------+-------------+-------------+-----------+
| 1            | 1           | 10          | 2020-06-10  | 1         |
| 2            | 1           | 20          | 2020-07-01  | 1         |
| 3            | 1           | 30          | 2020-07-08  | 2         |
| 4            | 2           | 10          | 2020-06-15  | 2         |
| 5            | 2           | 40          | 2020-07-01  | 10        |
| 6            | 3           | 20          | 2020-06-24  | 2         |
| 7            | 3           | 30          | 2020-06-25  | 2         |
| 9            | 3           | 30          | 2020-05-08  | 3         |
+--------------+-------------+-------------+-------------+-----------+

Result table:
+--------------+------------+
| customer_id  | name       |
+--------------+------------+
| 1            | Winston    |
+--------------+------------+
Winston spent \$300 (300 * 1) in June and \$100 ( 10 * 1 + 45 * 2) in July 2020.
Jonathan spent \$600 (300 * 2) in June and \$20 ( 2 * 10) in July 2020.
Moustafa spent \$110 (10 * 2 + 45 * 2) in June and \$0 in July 2020.
``````

MySQL Solution

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select customer_id, name
from
(
select o.customer_id, c.name,
sum(case when left(o.order_date,7) = '2020-06' then p.price * o.quantity end) as JuneSpend,
sum(case when left(o.order_date,7) = '2020-07' then p.price * o.quantity end) as JulySpend
from Orders o
left join Customers c on o.customer_id = c.customer_id
left join Product p on o.product_id = p.product_id
group by o.customer_id
having JuneSpend >= 100 and JulySpend >= 100
) as temp
``````