Given two rectangles, find if the given two rectangles overlap or not.
Note that a rectangle can be represented by two coordinates, top left and bottom right. So mainly we are given following four coordinates.
l1: Top Left coordinate of first rectangle.
r1: Bottom Right coordinate of first rectangle.
l2: Top Left coordinate of second rectangle.
r2: Bottom Right coordinate of second rectangle.

We need to write a function bool doOverlap(l1, r1, l2, r2) that returns true if the two given rectangles overlap.
Note : It may be assumed that the rectangles are parallel to the coordinate axis.
One solution is to one by one pick all points of one rectangle and see if the point lies inside the other rectangle or not. This can be done using the algorithm discussed here.
Following is a simpler approach. Two rectangles do not overlap if one of the following conditions is true.
1) One rectangle is above top edge of other rectangle.
2) One rectangle is on left side of left edge of other rectangle.
We need to check above cases to find out if given rectangles overlap or not. Following is the implementation of the above approach.
C++
#include <bits/stdc++.h>
struct Point {
int x, y;
};
bool doOverlap(Point l1, Point r1, Point l2, Point r2)
{
if (l1.x == r1.x || l1.y == r1.y || r2.x == l2.x || l2.y == r2.y)
return false ;
if (l1.x > r2.x || l2.x > r1.x)
return false ;
if (r1.y > l2.y || r2.y > l1.y)
return false ;
return true ;
}
int main()
{
Point l1 = { 0, 10 }, r1 = { 10, 0 };
Point l2 = { 5, 5 }, r2 = { 15, 0 };
if (doOverlap(l1, r1, l2, r2))
printf ( "Rectangles Overlap" );
else
printf ( "Rectangles Don't Overlap" );
return 0;
}
|
Java
class GFG {
static class Point {
int x, y;
}
static boolean doOverlap(Point l1, Point r1, Point l2, Point r2) {
if (l1.x == r1.x || l1.y == r1.y || r2.x == l2.x || l2.y == r2.y)
return false ;
if (l1.x > r2.x || l2.x > r1.x) {
return false ;
}
if (r1.y > l2.y || r2.y > l1.y) {
return false ;
}
return true ;
}
public static void main(String[] args) {
Point l1 = new Point(),r1 = new Point(),
l2 = new Point(),r2 = new Point();
l1.x= 0 ;l1.y= 10 ; r1.x= 10 ;r1.y= 0 ;
l2.x= 5 ;l2.y= 5 ; r2.x= 15 ;r2.y= 0 ;
if (doOverlap(l1, r1, l2, r2)) {
System.out.println( "Rectangles Overlap" );
} else {
System.out.println( "Rectangles Don't Overlap" );
}
}
}
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Python3
class Point:
def __init__( self , x, y):
self .x = x
self .y = y
def do_overlap(l1, r1, l2, r2):
if l1.x = = r1.x or l1.y = = r1.y or r2.x = = l2.x or l2.y = = r2.y:
return False
if l1.x > r2.x or l2.x > r1.x:
return False
if r1.y > l2.y or r2.y > l1.y:
return False
return True
if __name__ = = "__main__" :
l1 = Point( 0 , 10 )
r1 = Point( 10 , 0 )
l2 = Point( 5 , 5 )
r2 = Point( 15 , 0 )
if (do_overlap(l1, r1, l2, r2)):
print ( "Rectangles Overlap" )
else :
print ( "Rectangles Don't Overlap" )
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C#
using System;
class GFG
{
class Point
{
public int x, y;
}
static bool doOverlap(Point l1, Point r1,
Point l2, Point r2)
{
if (l1.x == r1.x || l1.y == r1.y || r2.x == l2.x || l2.y == r2.y)
{
return false ;
}
if (l1.x > r2.x || l2.x > r1.x)
{
return false ;
}
if (r1.y > l2.y || r2.y > l1.y)
{
return false ;
}
return true ;
}
public static void Main()
{
Point l1 = new Point(), r1 = new Point(),
l2 = new Point(), r2 = new Point();
l1.x = 0;l1.y = 10; r1.x = 10;r1.y = 0;
l2.x = 5;l2.y = 5; r2.x = 15;r2.y = 0;
if (doOverlap(l1, r1, l2, r2))
{
Console.WriteLine( "Rectangles Overlap" );
} else
{
Console.WriteLine( "Rectangles Don't Overlap" );
}
}
}
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Javascript
<script>
class Point {
constructor(val) {
this .x = val;
this .y = val;
}
}
function doOverlap( l1, r1, l2, r2) {
if (l1.x == r1.x || l1.y == r1.y || r2.x == l2.x || l2.y == r2.y)
return false ;
if (l1.x > r2.x || l2.x > r1.x) {
return false ;
}
if (r1.y > l2.y || r2.y > l1.y) {
return false ;
}
return true ;
}
var l1 = new Point(), r1 = new Point(),
l2 = new Point(), r2 = new Point();
l1.x = 0;
l1.y = 10;
r1.x = 10;
r1.y = 0;
l2.x = 5;
l2.y = 5;
r2.x = 15;
r2.y = 0;
if (doOverlap(l1, r1, l2, r2)) {
document.write( "Rectangles Overlap" );
} else {
document.write( "Rectangles Don't Overlap" );
}
</script>
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The Time Complexity of the above code is O(1) as the code doesn’t have any loop or recursion.
Auxiliary Space: O(1)
This article is compiled by Aman Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.