1571. Warehouse Manager

Level: Easy; Amazon

Question:

Table: Warehouse

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| name         | varchar |
| product_id   | int     |
| units        | int     |
+--------------+---------+
(name, product_id) is the primary key for this table.
Each row of this table contains the information of the products in each warehouse.

Table: Products

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| product_name  | varchar |
| Width         | int     |
| Length        | int     |
| Height        | int     |
+---------------+---------+
product_id is the primary key for this table.
Each row of this table contains information about the 
product dimensions (Width, Lenght, and Height) in feets of each product.

Write an SQL query to report the number of cubic feet of volume the inventory occupies in each warehouse.

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
Warehouse table:
+------------+--------------+-------------+
| name       | product_id   | units       |
+------------+--------------+-------------+
| LCHouse1   | 1            | 1           |
| LCHouse1   | 2            | 10          |
| LCHouse1   | 3            | 5           |
| LCHouse2   | 1            | 2           |
| LCHouse2   | 2            | 2           |
| LCHouse3   | 4            | 1           |
+------------+--------------+-------------+
Products table:
+------------+--------------+------------+----------+-----------+
| product_id | product_name | Width      | Length   | Height    |
+------------+--------------+------------+----------+-----------+
| 1          | LC-TV        | 5          | 50       | 40        |
| 2          | LC-KeyChain  | 5          | 5        | 5         |
| 3          | LC-Phone     | 2          | 10       | 10        |
| 4          | LC-T-Shirt   | 4          | 10       | 20        |
+------------+--------------+------------+----------+-----------+
Output:
 
+----------------+------------+
| warehouse_name | volume     | 
+----------------+------------+
| LCHouse1       | 12250      | 
| LCHouse2       | 20250      |
| LCHouse3       | 800        |
+----------------+------------+
Explanation:
 
Volume of product_id = 1 (LC-TV), 5x50x40 = 10000
Volume of product_id = 2 (LC-KeyChain), 5x5x5 = 125 
Volume of product_id = 3 (LC-Phone), 2x10x10 = 200
Volume of product_id = 4 (LC-T-Shirt), 4x10x20 = 800
LCHouse1: 1 unit of LC-TV + 10 units of LC-KeyChain + 5 units of LC-Phone.
          Total volume: 1*10000 + 10*125  + 5*200 = 12250 cubic feet
LCHouse2: 2 units of LC-TV + 2 units of LC-KeyChain.
          Total volume: 2*10000 + 2*125 = 20250 cubic feet
LCHouse3: 1 unit of LC-T-Shirt.
          Total volume: 1*800 = 800 cubic feet.

My Solution:

First: calculate the volume of each product (e.g. LC-TV, LC-KeyChain, etc.) and append it to each row in Warehouse with corresponding product_id.

Second: calculate the sum of the inventory using group by warehouse name.

select t.name as 'warehouse_name', sum(t.units*t.vol) as 'volume'
from (select name, w.product_id, Width*Length*Height as 'vol', units
      from Warehouse w inner join Products p on w.product_id=p.product_id) t
group by t.name