836. Rectangle Overlap.md

836. Rectangle Overlap

1. 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:

• rect1.length == 4
• rect2.length == 4
• -10^9 <= rec1[i], rec2[i] <= 10^9
• rec1 and rec2 represent a valid rectangle with a non-zero area.

2. Solutions

Solution 1: Language: C

• Wednesday, 20 April, 2022
• Time Complexity: $O(1)$
• Space Complexity: $O(1)$
• Runtime: 0 ms, faster than 100.00% of C online submissions for Rectangle Overlap.
• Memory Usage: 5.5 MB, less than 63.89% of C online submissions for Rectangle Overlap.

bool isRectangleOverlap(int *rec1, int rec1Size, int *rec2, int rec2Size) {
    // False conditions:
    // Notice the negation (!) below.
    return !(rec1[0] >= rec2[2] ||        // R1x1 >= R2x2
             rec1[2] <= rec2[0] ||        // R1x2 <= R2x1
             rec1[1] >= rec2[3] ||        // R1y1 >= R2y2
             rec1[3] <= rec2[1]);         // R1y2 <= R2y1
    // Otherwise True.
}

Solution 2: Language: Python Python Version of Solution 1

• Wednesday, 20 April, 2022
• Time Complexity: $O(1)$
• Space Complexity: $O(1)$
• Runtime: 39 ms, faster than 60.28% of Python3 online submissions for Rectangle Overlap.
• Memory Usage: 13.8 MB, less than 68.90% of Python3 online submissions for Rectangle Overlap.

class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        return not (rec1[0] >= rec2[2] or
                    rec1[2] <= rec2[0] or
                    rec1[1] >= rec2[3] or
                    rec1[3] <= rec2[1])

Solution 3: Language: Java Java Version of Solution 1

• Wednesday, 20 April, 2022
• Time Complexity: $O(1)$
• Space Complexity: $O(1)$
• Runtime: 0 ms, faster than 100.00% of Java online submissions for Rectangle Overlap.
• Memory Usage: 40.7 MB, less than 75.20% of Java online submissions for Rectangle Overlap.

class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return !(rec1[0] >= rec2[2] ||
                 rec1[2] <= rec2[0] ||
                 rec1[1] >= rec2[3] ||
                 rec1[3] <= rec2[1]);
    }
}