Mark and Toys Hacker Rank In Java | Java Solution

Hacker Rank Mark and Toys Solution In Java

Problem Statement :-

Mark and Jane are very happy after having their first child. Their son loves toys, so Mark wants to buy some. There are a number of different toys lying in front of him, tagged with their prices. Mark has only a certain amount to spend, and he wants to maximize the number of toys he buys with this money. Given a list of toy prices and an amount to spend, determine the maximum number of gifts he can buy.

Note : Each toy can be purchased only once. 

prices = [1,2,3,4]
k = 7

The budget is 7 units of currency. He can buy items that cost [1,2,3] for 6, or [3,4] for 7 units. The maximum is 3 items.

Example 1 :

Input :
prices = [1, 2, 3, 4]
k = 7

Output :
3

Explanation :
The budget is 7 units of currency. He can buy items that cost [1,2,3] for 6, or [3,4] for 7 units. The maximum is 3 items.

Example 2 :

Input :
prices = [20, 3, 10, 8, 1, 12, 5, 6]
k = 30

Output :
5

Explanation :
The budget is 30 units of currency. He can buy items that cost [1, 3, 5, 6, 8] for 23. The maximum is 5 items Mark can buy.

See description on Hacker Rank :-

• https://www.hackerrank.com/challenges/mark-and-toys/problem 
  

So now lets see Solution of above problem.

For solution, first we have to Sort the array so it becomes easy to find maximum number of toys Mark can buy. 

We traverse through entire array and doing sum of numbers until we reach given k currency. Total sum can not be greater than given k currency. 

Solution 1 : Java code for Mark and Toys Problem

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    // Complete the maximumToys function below.
    static int maximumToys(int[] priceArray, int price) {
       
        Arrays.sort(priceArray);
        
        int totalToys = 0;
        int totalToysPrice = 0;
        
        for (int i = 0; i < priceArray.length-1; i++) {
            if (price >= totalToysPrice) {
                totalToysPrice += priceArray[i];
                if (price >= totalToysPrice) {
                    totalToys++;
                }
            }
        }
        return totalToys;
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new
                FileWriter(System.getenv("OUTPUT_PATH")));

        String[] nk = scanner.nextLine().split(" ");

        int n = Integer.parseInt(nk[0]);

        int k = Integer.parseInt(nk[1]);

        int[] prices = new int[n];

        String[] pricesItems = scanner.nextLine().split(" ");
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        for (int i = 0; i < n; i++) {
            int pricesItem = Integer.parseInt(pricesItems[i]);
            prices[i] = pricesItem;
        }

        int result = maximumToys(prices, k);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedWriter.close();

        scanner.close();
    }
}

Explanation :-

1. Sort the array using Arrays.sort(). so in array lesser value comes first and greater value last.
   
2. Loop through Given prices array and sum the array value 1 by 1 into totalToys variable.
   
3. Check if price is still less than sum of array values.
4. Return totalToys.

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