warehouse-manager.sql

/*

Warehouse Manager Problem

Description
LeetCode Problem 1571.

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 the information about the product dimensions (Width, Lenght and Height) in feets of each product.
Write an SQL query to report how much cubic feet of volume does the inventory occupy in each warehouse.

warehouse_name
volume Return the result table in any order.
The query result format is in the following example.

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        |
+------------+--------------+------------+----------+-----------+

Result table:
+----------------+------------+
| warehouse_name | volume     | 
+----------------+------------+
| LCHouse1       | 12250      | 
| LCHouse2       | 20250      |
| LCHouse3       | 800        |
+----------------+------------+
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.

*/


# V0
select name as warehouse_name, sum(units * vol) as volume
from Warehouse w
join (select product_id, Width*Length*Height as vol
     from Products) p
on w.product_id = p.product_id
group by name

# V1
# https://circlecoder.com/warehouse-manager/
select name as warehouse_name, sum(units * vol) as volume
from Warehouse w
join (select product_id, Width*Length*Height as vol
     from Products) p
on w.product_id = p.product_id
group by name

# V2
# Time:  O(n)
# Space: O(n)
SELECT name AS warehouse_name,
       SUM(units*Width*LENGTH*Height) AS volume
FROM Warehouse w
INNER JOIN Products p ON w.product_id = p.product_id
GROUP BY name
ORDER BY NULL;