836. Rectangle Overlap LeetCode Solution

In this guide, you will get 836. Rectangle Overlap LeetCode Solution with the best time and space complexity. The solution to Rectangle Overlap problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Rectangle Overlap solution in C++
Rectangle Overlap solution in Java
Rectangle Overlap solution in Python
Additional Resources

836. Rectangle Overlap LeetCode Solution image

Problem Statement of Rectangle Overlap

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.

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(1)

836. Rectangle Overlap LeetCode Solution in C++

class Solution {
 public:
  bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
    return rec1[0] < rec2[2] && rec2[0] < rec1[2] &&  //
           rec1[1] < rec2[3] && rec2[1] < rec1[3];
  }
};
/* code provided by PROGIEZ */

836. Rectangle Overlap LeetCode Solution in Java

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];
  }
}
// code provided by PROGIEZ

836. Rectangle Overlap LeetCode Solution in Python

class Solution:
  def isRectangleOverlap(self, rec1: list[int], rec2: list[int]) -> bool:
    return rec1[0] < rec2[2] and rec2[0] < rec1[2] and rec1[1] < rec2[3] and rec2[1] < rec1[3]
 # code by PROGIEZ