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 == 4rec2.length == 4-109 <= rec1[i], rec2[i] <= 109rec1andrec2represent 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]));
}
};
Input
Output
#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]));
}
}
Input
Output
#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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Output
#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
Input
Output
#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];
}
}
}
Input
Output