Formatted question description: https://leetcode.ca/all/1571.html

1571. Warehouse Manager

Level

Easy

Description

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.

Solution

The volume of each product is calculated as units * Width * Length * Height. For each warehouse, join tables Warehouse and Products using the field product_id and calculate the sum of all products’ volumes. Finally, group the result by warehouse_name.

# Write your MySQL query statement below
select warehouse_name, sum(volume) as volume from (
    select w.name as warehouse_name, w.product_id, w.units * Width * Length * Height as volume
    from Warehouse w left join Products p on w.product_id = p.product_id
) t group by warehouse_name;

Formatted question description: https://leetcode.ca/all/1571.html

All Problems

All Solutions