leetcode/solution/0800-0899/0836.Rectangle Overlap

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0836.Rectangle%20Overlap/README_EN.md	Geometry Math

836. Rectangle Overlap

中文文档

Description

An axis-aligned rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis.

Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two axis-aligned rectangles rec1 and rec2, return true if they overlap, otherwise return false.

 

Example 1:

Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
Output: true

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
Output: false

Example 3:

Input: rec1 = [0,0,1,1], rec2 = [2,2,3,3]
Output: false

 

Constraints:

• rec1.length == 4
• rec2.length == 4
• -109 <= rec1[i], rec2[i] <= 109
• rec1 and rec2 represent a valid rectangle with a non-zero area.

Solutions

Solution 1: Determine Non-Overlap Cases

Let the coordinates of rectangle \text{rec1} be (x_1, y_1, x_2, y_2), and the coordinates of rectangle \text{rec2} be (x_3, y_3, x_4, y_4).

The rectangles \text{rec1} and \text{rec2} do not overlap if any of the following conditions are met:

• y_3 \geq y_2: \text{rec2} is above \text{rec1};
• y_4 \leq y_1: \text{rec2} is below \text{rec1};
• x_3 \geq x_2: \text{rec2} is to the right of \text{rec1};
• x_4 \leq x_1: \text{rec2} is to the left of \text{rec1}.

If none of the above conditions are met, the rectangles \text{rec1} and \text{rec2} overlap.

The time complexity is O(1), and the space complexity is O(1).

Python3

class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        x1, y1, x2, y2 = rec1
        x3, y3, x4, y4 = rec2
        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)

Java

class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
}

C++

class Solution {
public:
    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
};

Go

func isRectangleOverlap(rec1 []int, rec2 []int) bool {
	x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
	x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
	return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
}

TypeScript

function isRectangleOverlap(rec1: number[], rec2: number[]): boolean {
    const [x1, y1, x2, y2] = rec1;
    const [x3, y3, x4, y4] = rec2;
    return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
}