README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1459.Rectangles%20Area/README_EN.md	Database

1459. Rectangles Area 🔒

中文文档

Description

Table: Points

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| x_value       | int     |
| y_value       | int     |
+---------------+---------+
id is the column with unique values for this table.
Each point is represented as a 2D coordinate (x_value, y_value).

 

Write a solution to report all possible axis-aligned rectangles with a non-zero area that can be formed by any two points from the Points table.

Each row in the result should contain three columns (p1, p2, area) where:

• p1 and p2 are the id's of the two points that determine the opposite corners of a rectangle.
• area is the area of the rectangle and must be non-zero.

Return the result table ordered by area in descending order. If there is a tie, order them by p1 in ascending order. If there is still a tie, order them by p2 in ascending order.

The result format is in the following table.

 

Example 1:

Input: 
Points table:
+----------+-------------+-------------+
| id       | x_value     | y_value     |
+----------+-------------+-------------+
| 1        | 2           | 7           |
| 2        | 4           | 8           |
| 3        | 2           | 10          |
+----------+-------------+-------------+
Output: 
+----------+-------------+-------------+
| p1       | p2          | area        |
+----------+-------------+-------------+
| 2        | 3           | 4           |
| 1        | 2           | 2           |
+----------+-------------+-------------+
Explanation: 
The rectangle formed by p1 = 2 and p2 = 3 has an area equal to |4-2| * |8-10| = 4.
The rectangle formed by p1 = 1 and p2 = 2 has an area equal to |2-4| * |7-8| = 2.
Note that the rectangle formed by p1 = 1 and p2 = 3 is invalid because the area is 0.

Solutions

Solution 1

MySQL

# Write your MySQL query statement below
SELECT
    p1.id AS p1,
    p2.id AS p2,
    ABS(p1.x_value - p2.x_value) * ABS(p1.y_value - p2.y_value) AS area
FROM
    Points AS p1
    JOIN Points AS p2 ON p1.id < p2.id
WHERE p1.x_value != p2.x_value AND p1.y_value != p2.y_value
ORDER BY area DESC, p1, p2;