836. Rectangle Overlap

Problem 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.

Examples:

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

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

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

Python Solution

class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        # Check if one rectangle is to the left of the other
        if rec1[2] <= rec2[0] or rec2[2] <= rec1[0]:
            return False
        
        # Check if one rectangle is above the other
        if rec1[3] <= rec2[1] or rec2[3] <= rec1[1]:
            return False
        
        return True

Implementation Notes:

• Uses simple coordinate comparison
• Checks for non-overlap conditions
• Time complexity: O(1)
• Space complexity: O(1)

Java Solution

class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        // Check if one rectangle is to the left of the other
        if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {
            return false;
        }
        
        // Check if one rectangle is above the other
        if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {
            return false;
        }
        
        return true;
    }
}

C++ Solution

class Solution {
public:
    bool isRectangleOverlap(vector& rec1, vector& rec2) {
        // Check if one rectangle is to the left of the other
        if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {
            return false;
        }
        
        // Check if one rectangle is above the other
        if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {
            return false;
        }
        
        return true;
    }
};

JavaScript Solution

/**
 * @param {number[]} rec1
 * @param {number[]} rec2
 * @return {boolean}
 */
var isRectangleOverlap = function(rec1, rec2) {
    // Check if one rectangle is to the left of the other
    if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {
        return false;
    }
    
    // Check if one rectangle is above the other
    if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {
        return false;
    }
    
    return true;
};

C# Solution

public class Solution {
    public bool IsRectangleOverlap(int[] rec1, int[] rec2) {
        // Check if one rectangle is to the left of the other
        if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {
            return false;
        }
        
        // Check if one rectangle is above the other
        if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {
            return false;
        }
        
        return true;
    }
}

Implementation Notes:

• Simple and efficient implementation
• Uses logical negation of non-overlap conditions
• Time complexity: O(1)
• Space complexity: O(1)