Java Solution to Codility’s Number Of Disc Intersections Problem

 
import java.util.Arrays;
 
class Solution {
    public int solution(int[] A) {
    	// variable holding max intersections given in problem description (10,000,000)
    	final int MAX_INTERSECTIONS = 10000000;
    	// variable to keep the length of A
    	int N = A.length;
    	//initialize arrays to store our discs' end points
        // left end points
    	int [] discStartPoint = new int [N];
        // right end points
    	int [] discEndPoint = new int [N];		
 
    	//populate the arrays
    	for (int i = 0; i < N; i++) {
    		discStartPoint[i] = i - A[i];
    		// if we can overflow, Example: i = 5; A[i] = Integer.MAX_VALUE; discEndPoint[i] = 5 + Integer.MAX_VALUE, causes overflow
    		if (Integer.MAX_VALUE - A[i] < i) {
    			discEndPoint[i] = Integer.MAX_VALUE;    			
    		}
    		else {
    			discEndPoint[i] = i + A[i];
    		}
    	}
 
    	// sort the arrays
    	Arrays.sort(discStartPoint);
    	Arrays.sort(discEndPoint);
 
    	//variable for traversing "discStartPoint" array
    	int startPointIndex= 0;
    	//variable for traversing "discEndPoint" array 
    	int endPointIndex = 0;
 
    	// How many discs are currently open 
    	int openDiscs = 0;
    	// number of intersections
    	int intersections = 0;
 
    	//While we have not went through all of the discs available
    	//loop ends when all discs are opened
    	while(startPointIndex < N) {
    		// if we open a disc
    		// every time we open a disc, the disc will intersect with all previously opened discs
    		// and this new disc will be used for calculating future intersections
    		if (discStartPoint[startPointIndex] <= discEndPoint[endPointIndex]) {
                        intersections = intersections + openDiscs; 
                        // if our intersections are greater than max_intersections (see problem description) 
                        if (intersections > MAX_INTERSECTIONS) {
                	    return -1;
                        }
                        openDiscs++;
                        startPointIndex++;
    		}
    		// else, we close a disc
    		// this means that we have one less disc to intersect with
    		else {
    			openDiscs--;
                        endPointIndex++;
                }
 
    	}
 
    	return intersections;
    }
}