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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0836_rectangle_overlap;

// #Easy #Math #Geometry #2022_03_24_Time_0_ms_(100.00%)_Space_41.6_MB_(35.43%)

/**
 * 836 - Rectangle Overlap\.
 *
 * Easy
 *
 * 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`
 * *   -10⁹ <= rec1[i], rec2[i] <= 10⁹
 * *   `rec1` and `rec2` represent a valid rectangle with a non-zero area.
**/
public class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return rec1[0] < rec2[2] && rec2[0] < rec1[2] && rec1[1] < rec2[3] && rec2[1] < rec1[3];
    }
}