# Maximise the number of toys that can be purchased with amount K using min Heap

Given an array **arr[]** consisting of the cost of toys and an integer **K** depicting the amount of money available to purchase toys. The task is 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:arr[] = {1, 12, 5, 111, 200, 1000, 10, 9, 12, 15}, K = 50

Output:6

Toys with amount 1, 5, 9, 10, 12, and 12

can be purchased resulting in a total amount of 49.

Hence, the maximum number of toys are 6.

Input:arr[] = {1, 12, 5, 111, 200, 1000, 10}, K = 50

Output:4

**Approach:** Insert all the elements of the given array in a priority_queue now one by one remove elements from this priority queue and add these costs in a variable **sum** initialised to **0**. Keep removing the elements while the new addition keep the sum smaller than **K**. In the end, the count of elements removed will be the required answer.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the count of ` `// maximum toys that can be bought ` `int` `maxToys(` `int` `arr[], ` `int` `n, ` `int` `k) ` `{ ` ` ` ` ` `// Create a priority_queue and push ` ` ` `// all the array elements in it ` ` ` `priority_queue<` `int` `, vector<` `int` `>, greater<` `int` `> > pq; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `pq.push(arr[i]); ` ` ` `} ` ` ` ` ` `// To store the count of maximum ` ` ` `// toys that can be bought ` ` ` `int` `count = 0; ` ` ` `while` `(pq.top() <= k) { ` ` ` `count++; ` ` ` `k = k - pq.top(); ` ` ` `pq.pop(); ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 12, 5, 111, 200, 1000, 10 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `int` `k = 50; ` ` ` ` ` `cout << maxToys(arr, n, k); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

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**Output:**

4

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