Fiberlink online coding challenge - Program | Railway ticket counter and revenue problem

gistfile1.java

import java.util.Arrays;
import java.util.Scanner;

public class RailwayTickets {
	int numOfTickets, ticketsToSell;
	int[] ticketsPerBooth;
	
	public RailwayTickets(int numOfTickets, int ticketsToSell, int[] ticketsPerBooth)
	{
		this.numOfTickets = numOfTickets;
		this.ticketsToSell = ticketsToSell;
		this.ticketsPerBooth = ticketsPerBooth;
	}
	
	public int findMaxRevenue()
	{
		int[] perBooth = this.ticketsPerBooth;
		Arrays.sort(perBooth);
		int i = perBooth.length-1;
		int revenue = 0;
		while(i >= 0 && this.ticketsToSell != 0)
		{
			int nxtBooth = 0;
			int currBooth = perBooth[i];
			if(i == 0)
			{
				nxtBooth = 0;
			}
			else
			{
				nxtBooth = perBooth[i-1];
			}
			
			int diff = currBooth - nxtBooth;
			while(diff != 0 && this.ticketsToSell != 0)
			{
				revenue += currBooth;
				currBooth--;
				diff--;
				this.ticketsToSell--;
			}
			i--;
		}
		return revenue;
	}
	
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int numOfBooths = sc.nextInt();
		int ticketsToSell = sc.nextInt();
		int perBooth[] = new int[numOfBooths];
		for(int i = 0; i < numOfBooths; i++)
		{
			perBooth[i] = sc.nextInt();
		}
		RailwayTickets rt = new RailwayTickets(numOfBooths, ticketsToSell, perBooth);
		System.out.println("Max Revenue = " + rt.findMaxRevenue());
	}
}

There are "n" ticket windows in the railway station. i'th window has Ai tickets available. Price of a ticket is equal to the number of tickets remaining in that window at that time. When "m" tickets have been sold, what's the maximum amount of money the railway station can earn?
Ex. n=2, m=4
in 2 window available tickets are : 2 , 5
2nd window sold 4 tickets so 5+4+3+2=14.