g0201_0300.s0223_rectangle_area.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0201_0300.s0223_rectangle_area;

// #Medium #Math #Geometry #2022_07_04_Time_4_ms_(65.35%)_Space_43.4_MB_(31.18%)

/**
 * 223 - Rectangle Area\.
 *
 * Medium
 *
 * Given the coordinates of two **rectilinear** rectangles in a 2D plane, return _the total area covered by the two rectangles_.
 *
 * The first rectangle is defined by its **bottom-left** corner `(ax1, ay1)` and its **top-right** corner `(ax2, ay2)`.
 *
 * The second rectangle is defined by its **bottom-left** corner `(bx1, by1)` and its **top-right** corner `(bx2, by2)`.
 *
 * **Example 1:**
 *
 * ![Rectangle Area](https://assets.leetcode.com/uploads/2021/05/08/rectangle-plane.png)
 *
 * **Input:** ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2
 *
 * **Output:** 45 
 *
 * **Example 2:**
 *
 * **Input:** ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2
 *
 * **Output:** 16 
 *
 * **Constraints:**
 *
 * *   -10⁴ <= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 <= 10⁴
**/
@SuppressWarnings("java:S107")
public class Solution {
    public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        long left = Math.max(ax1, bx1);
        long right = Math.min(ax2, bx2);
        long top = Math.min(ay2, by2);
        long bottom = Math.max(ay1, by1);
        long area = (right - left) * (top - bottom);
        // if not overlaping, either of these two will be non-posittive
        // if right - left = 0, are will automtically be 0 as well
        if (right - left < 0 || top - bottom < 0) {
            area = 0;
        }
        return (int) ((ax2 - ax1) * (ay2 - ay1) + (bx2 - bx1) * (by2 - by1) - area);
    }
}