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Maximise the number of toys that can be purchased with amount K
• Difficulty Level : Easy
• Last Updated : 10 Jul, 2019

Given an array consisting of 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 = 50
cost = { 1, 12, 5, 111, 200, 1000, 10 }
Output: 4
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 the 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 equals to K.

Below image 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 ``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);`` ` `    ``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

 ``
Output :
```6
```

Time Complexity : O(N * logN), where N is the size of cost array.

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