Algorithm : 

  • Two rectangles can overlap if one rectangle has 0,1,2,4 corners inside the other rectangle. The check of above mentioned condition could result is many different combinations. Remember overlapping rectangle cannot have 3 corners inside.
  • Other way we can say that two rectangle overlap if the region of one rectangle lies inside the other.

various overlapping possibilites

  • The best way to find is to identify whether an overlapping area is present or not which can be known if the below mentioned all conditions are true.

If we check that

·        The left edge of B is to the left of right edge of R.

·         The top edge of B is above the R bottom edge.

·         The right edge of B is to the right of left edge of R.

·        The bottom edge of B is below the R upper edge.

Then we can say that rectangles are overlapping.


Consider an example: There are two rectangles as shown in diagram – Black Rectangle (B) and Red rectangle(R).



If we check that

·        The left edge of B  is to the left of  right edge of  R. The selected area will be :




·        The top edge of B is above the R bottom edge. So the selected area will be:


·        The right edge of B is to the right of left edge of R. The selected area will be:






·        The bottom edge of B is below the R upper edge. The selected area will be:




Hence all conditions are true we can say that rectangles are overlapping.

Therefore we can see that all the conditions are valid and hence rectangle is overlapping.


Below is the source code attached. The logic is only present in checkOverlap() method, rest is used to display the GUI and check.


package com;


import java.awt.*;

import java.awt.event.*;


import javax.swing.*;


public class RectangleOverlap extends JPanel{


      int r1x1,r1x2,r2x1,r2x2 ;

      int r1y1,r1y2,r2y1,r2y2 ;

            int r1width,r2width ;

            int r1height,r2height ;

      static JButton btn = new JButton(“Check”);

      public RectangleOverlap(int r1x1,int r1y1,int r1x2,int r1y2,int r2x1,int r2y1,int r2x2,int r2y2){









            r1width = Math.abs(r1x1-r1x2);

            r2width = Math.abs(r2x1-r2x2);

            r1height = Math.abs(r1y1-r1y2);

            r2height = Math.abs(r2y1-r2y2);




      private void addActionListener() {

            btn.addActionListener(new ActionListener(){


                  public void actionPerformed(ActionEvent e) {






      private void checkOverlap() {

            // condition to check whether the rectangles are overlapping or not.s

            boolean isOVerlap= ((r1x2 >= r2x1) &&

                  (r1y2 >= r2y1) &&

                  (r1x1 <= r2x2) &&

                (r1y1 <= r2y2));


      if(isOVerlap ){

            JOptionPane.showMessageDialog(null, “OVerlap”);


            JOptionPane.showMessageDialog(null, “No OVerlap”);





      protected void paintComponent(Graphics g) {

            g.drawRect(r1x1,r1y1 , r1width, r1height);

            g.setColor(new Color(123,232,122));

            g.drawRect(r2x1, r2y1, r2width,r2height);



      public static void main(String args[]){

            JFrame frame = new JFrame();


// input to check overlap condition.

// the order followed as : enter coordinate for 1st rectangle as lower limit and upper limit to rectangles          

            frame.getContentPane().add(new RectangleOverlap(20,30,120,130,10,50,160,120),BorderLayout.CENTER);


            frame.addWindowListener(new WindowAdapter(){


                  public void windowClosing(WindowEvent e) {