/**

* Created by Vatsal Gosaliya on 15-Jul-16.

*

* PROBLEM: Given are two rectangles defined over the 2-D cartesian coordinate system.

* Each rectangle is expressed using the top-left corner, width and height.

* The aim is to detect whether they overlap, and compute the overlapping

* area, if they do.

*/

class OverlappingRectangles {

private double x,y,width,height;

OverlappingRectangles(double x, double y, double width, double height){

this.x = x;

this.y = y;

this.width = width;

this.height = height;

}

OverlappingRectangles intersection(OverlappingRectangles rect2) {

double newX = Math.max(this.x, rect2.x);

double newY = Math.min(this.y, rect2.y);

//System.out.println("newX = "+newX);

//System.out.println("newY = "+newY);

double newWidth = Math.min(this.x + this.width, rect2.x + rect2.width) - newX;

double newHeight = newY - Math.max(this.y - this.height, rect2.y - rect2.height);

//System.out.println("newWidth = "+newWidth);

//System.out.println("newHeight = "+newHeight);

if (newWidth <= 0d || newHeight <= 0d){

System.out.print("No intersection.");

return null;

}

return new OverlappingRectangles(newX, newY, newWidth, newHeight);

}

public double getArea(){

return this.width*this.height;

}

public static void main(String args[]){

OverlappingRectangles r1 = new OverlappingRectangles(2,8,3,4);

OverlappingRectangles r2 = new OverlappingRectangles(1,7,6,2);

OverlappingRectangles r = r1.intersection(r2);

double areaOfIntersection = r == null ? 0 : r.getArea();

System.out.print("Area of intersection : "+areaOfIntersection);

}

}