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

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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