Maximise the number of toys that can be purchased with amount K

Given an array consisting of the cost of toys. Given an integer K depicting the amount of money available to purchase toys. Write a program to find the maximum number of toys one can buy with the amount K. 

Note: One can buy only 1 quantity of a particular toy.

Examples:

Input:  N = 10, K =  50,  cost = { 1, 12, 5, 111, 200, 1000, 10, 9, 12, 15 }
Output: 6
Explanation: Toys with amount 1, 5, 9, 10, 12, and 12  can be purchased resulting in a total amount of 49. Hence, maximum number of toys is 6.

Input: N = 7, K = 5,  cost = { 1, 12, 5, 111, 200, 1000, 10 }
Output: 4 

The idea to solve this problem is to first sort the cost array in ascending order. This will arrange the toys in increasing order of cost. Now iterate over the cost array and keep calculating the sum of costs until the sum is less than or equal to K. Finally, return the number of toys used to calculate the sum which is just less than or equal to K.

The image below is an illustration of the above approach:  

Below is the implementation of the above approach: 

C++

// C++ Program to maximize the
// number of toys with K amount
#include <bits/stdc++.h>
using namespace std;

// This functions returns the required
// number of toys
int maximum_toys(int cost[], int N, int K)
{
    int count = 0, sum = 0;

    // sort the cost array
    sort(cost, cost + N);
    for (int i = 0; i < N; i++) {

        // Check if we can buy ith toy or not
        if (sum +cost[i] <= K)
        {
            sum = sum + cost[i];
            // Increment count
            count++;
        }
    }
    return count;
}

// Driver Code
int main()
{
    int K = 50;
    int cost[] = { 1, 12, 5, 111, 200, 1000, 10, 9, 12, 15 };
    int N = sizeof(cost) / sizeof(cost[0]);

    cout << maximum_toys(cost, N, K) << endl;
    return 0;
}

Java

// Java Program to maximize the
// number of toys with K amount
import java.io.*;
import java .util.*;

class GFG
{
// This functions returns
// the required number of toys
static int maximum_toys(int cost[],
                        int N, int K)
{
    int count = 0, sum = 0;

    // sort the cost array
    Arrays.sort(cost);
    for (int i = 0; i < N; i++)
    {

        // Check if we can buy ith toy or not
        if (sum +cost[i] <= K)
        {
            sum = sum + cost[i];
            // Increment count
            count++;
        }
    }
    return count;
}

// Driver Code
public static void main (String[] args)
{
int K = 50;
int cost[] = {1, 12, 5, 111, 200,
            1000, 10, 9, 12, 15};
int N = cost.length;

System.out.print( maximum_toys(cost, N, K));
}
}

// This code is contributed by anuj_67.

Python3

# Python 3 Program to maximize the
# number of toys with K amount

# This functions returns the required
# number of toys
def maximum_toys(cost, N, K):
    count = 0
    sum = 0

    # sort the cost array
    cost.sort(reverse = False)
    for i in range(0, N, 1):

        # Check if we can buy ith toy or not
        if (sum+cost[i] <= K):
            sum = sum + cost[i]
            # Increment the count variable
            count += 1

    return count

# Driver Code
if __name__ == '__main__':
    K = 50
    cost = [1, 12, 5, 111, 200,
            1000, 10, 9, 12, 15]
    N = len(cost)

    print(maximum_toys(cost, N, K))

# This code is contributed by
# Sanjit_Prasad

C#

// C# Program to maximize the
// number of toys with K amount
using System;

class GFG
{
// This functions returns
// the required number of toys
static int maximum_toys(int []cost,
                        int N, int K)
{
    int count = 0, sum = 0;

    // sort the cost array
    Array.Sort(cost);
    for (int i = 0; i < N; i++)
    {

        // Check if we can buy ith toy or not
        if (sum +cost[i] <= K)
        {
            sum = sum + cost[i];
            // Increment count
            count++;
        }
    }
    return count;
}

// Driver Code
public static void Main ()
{
int K = 50;
int []cost = {1, 12, 5, 111, 200,
            1000, 10, 9, 12, 15};
int N = cost.Length;

Console.Write( maximum_toys(cost, N, K));
}
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP Program to maximize the
// number of toys with K amount

// This functions returns
// the required number of toys
function maximum_toys($cost, $N, $K)
{
    $count = 0; $sum = 0;

    // sort the cost array
        sort($cost);
    for ($i = 0; $i < $N; $i++)
    {

            // Check if we can buy ith toy or not
        if ($sum + $cost[$i] <= $K)
        {
            $sum = $sum + $cost[$i];
            // Increment the count variable
            $count++;
        }
    }
    return $count;
}

// Driver Code
$K = 50;
$cost = array(1, 12, 5, 111, 200,
            1000, 10, 9, 12, 15 );
$N = count($cost);

echo maximum_toys($cost, $N, $K),"\n";

// This code is contributed by anuj_67
?>

Javascript

<script>
    // Javascript Program to maximize the
    // number of toys with K amount

    // This functions returns
    // the required number of toys
    function maximum_toys(cost, N, K)
    {
        let count = 0, sum = 0;

        // sort the cost array
        cost.sort(function(a, b){return a - b});
        for (let i = 0; i < N; i++)
        {

            // Check if we can buy ith toy or not
            if (sum +cost[i] <= K)
            {
                sum = sum + cost[i];
                // Increment count
                count++;
            }
        }
        return count;
    }

    let K = 50;
    let cost = [1, 12, 5, 111, 200, 1000, 10, 9, 12, 15];
    let N = cost.length;

    document.write(maximum_toys(cost, N, K));

</script>

Output

6

Time Complexity: O(N * logN), where N is the size of the cost array.
Auxiliary Space: O(1) as it is using constant space for variables