Find if two rectangles overlap

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    struct Point {     int x, y; };    // Returns true if two rectangles (l1, r1) and (l2, r2) overlap bool doOverlap(Point l1, Point r1, Point l2, Point r2) {     // If one rectangle is on left side of other     if (l1.x > r2.x || l2.x > r1.x)         return false;        // If one rectangle is above other     if (l1.y < r2.y || l2.y < r1.y)         return false;        return true; }    /* Driver program to test above function */ 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

 // Java programs to search a word in a 2D grid    class GFG {       static class Point {            int x, y;     }    // Returns true if two rectangles (l1, r1) and (l2, r2) overlap   static  boolean doOverlap(Point l1, Point r1, Point l2, Point r2) {         // If one rectangle is on left side of other          if (l1.x > r2.x || l2.x > r1.x) {             return false;         }            // If one rectangle is above other          if (l1.y < r2.y || l2.y < r1.y) {             return false;         }            return true;     }        /* Driver program to test above function */     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");         }     } } //this code contributed by PrinciRaj1992

C#

 // C# programs to search a word in a 2D grid  using System;        class GFG  {     class Point      {         public int x, y;     }        // Returns true if two rectangles (l1, r1)      // and (l2, r2) overlap      static bool doOverlap(Point l1, Point r1,                           Point l2, Point r2)      {         // If one rectangle is on left side of other          if (l1.x > r2.x || l2.x > r1.x)         {             return false;         }            // If one rectangle is above other          if (l1.y < r2.y || l2.y < r1.y)          {             return false;         }         return true;     }        // Driver Code     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. = 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");         }     } }    // This code is contributed by  // Rajput-Ji

Output:

Rectangles Overlap

Time Complexity of above code is O(1) as the code doesn’t have any loop or recursion.