Algorithm


Problem Name: 836. 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.
 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming


class Solution {
public:
    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
        return (min(rec1[3], rec2[3]) > max(rec1[1], rec2[1])) && (min(rec1[2], rec2[2]) > max(rec1[0], rec2[0]));
    }
};
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Input

x
+
cmd
rec1 = [0,0,2,2], rec2 = [1,1,3,3]

Output

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true

#2 Code Example with Java Programming

Code - Java Programming


class Solution {
  public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
    return (Math.min(rec1[2], rec2[2]) > Math.max(rec1[0], rec2[0]) && 
            Math.min(rec1[3], rec2[3]) > Math.max(rec1[1], rec2[1])); 
  }
}
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Input

x
+
cmd
rec1 = [0,0,2,2], rec2 = [1,1,3,3]

Output

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true

#3 Code Example with Javascript Programming

Code - Javascript Programming


const isRectangleOverlap = function(rec1, rec2) {
  return !(
    chkOverlap(rec1, rec2) === false || chkOverlap(rec2, rec1) === false
  );
};
function chkOverlap(r1, r2) {
  if (r1[2]  < = r2[0] || r1[3] <= r2[1]) {
    return false;
  } else {
    return true;
  }
}

console.log(isRectangleOverlap([0, 0, 2, 2], [1, 1, 3, 3]));
console.log(isRectangleOverlap([0, 0, 1, 1], [1, 0, 2, 1]));
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Input

x
+
cmd
rec1 = [0,0,1,1], rec2 = [1,0,2,1]

Output

x
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false

#4 Code Example with Python Programming

Code - Python Programming


class Solution:
    def isRectangleOverlap(self, rec1, rec2):
        x = (rec1[2] - rec1[0] + rec2[2] - rec2[0]) > (max(rec1[2], rec2[2]) - min(rec1[0], rec2[0]))
        y = (rec1[3] - rec1[1] + rec2[3] - rec2[1]) > (max(rec1[3], rec2[3]) - min(rec1[1], rec2[1]))
        return x and y
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Input

x
+
cmd
rec1 = [0,0,1,1], rec2 = [1,0,2,1]

Output

x
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false

#5 Code Example with C# Programming

Code - C# Programming


namespace LeetCode
{
    public class _0836_RectangleOverlap
    {
        public bool IsRectangleOverlap(int[] rec1, int[] rec2)
        {
            return rec1[2] > rec2[0] &&
                   rec1[3] > rec2[1] &&
                   rec1[0] < rec2[2] &&
                   rec1[1] < rec2[3];
        }
    }
}
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Input

x
+
cmd
rec1 = [0,0,1,1], rec2 = [2,2,3,3]

Output

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cmd
false
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