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

• Difficulty Level : Easy
• Last Updated : 05 May, 2022

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:
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++

 `// C++ implementation of the approach``#include ``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);``    ``int` `k = 50;` `    ``cout << maxToys(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{``    ` `// Function to return the count of``// maximum toys that can be bought    ``public` `static` `int` `maxToys(``int``[] arr, ``int` `k)``{``    ``int` `n = arr.length;``    ` `    ``// Create a priority_queue and push``    ``// all the array elements in it``    ``PriorityQueue pq = ``new` `PriorityQueue();``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``pq.offer(arr[i]);``    ``}``    ` `    ``// To store the count of maximum``    ``// toys that can be bought``    ``int` `count = ``0``;``    ``while` `(!pq.isEmpty() && pq.peek() <= k)``    ``{``        ``k = k - pq.poll();``        ``count++;``    ``}``    ``return` `count;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int``[] arr = ``new` `int``[]{ ``1``, ``12``, ``5``, ``111``,``                           ``200``, ``1000``, ``10` `};``    ``int` `k = ``50``;                      ``    ` `      ``System.out.println(maxToys(arr, k));     ``}``}` `// This code is contributed by ankit bajpai`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# maximum toys that can be bought` `# importing heapq module``import` `heapq`  `def` `maxToys(arr, n, k):``    ``# Create a priority_queue and push``    ``# all the array elements in it``    ``pq ``=` `arr``    ``heapq.heapify(pq)``    ``# To store the count of maximum``    ``# toys that can be bought``    ``count ``=` `0` `    ``while` `(pq[``0``] <``=` `k):``        ``count ``+``=` `1``        ``k ``-``=` `pq[``0``]``        ``# assigning last element of the min heap``        ``# to top of the heap``        ``pq[``0``] ``=` `pq[``-``1``]``        ``# deleting the last element.``        ``pq.pop()``        ``# pq.pop() is an O(1) operation``        ``# maintaining the heap property again` `        ``heapq.heapify(pq)``    ``return` `count`  `# Driver code``arr ``=` `[``1``, ``12``, ``5``, ``111``, ``200``, ``1000``, ``10``]``n ``=` `len``(arr)``k ``=` `50``print``(maxToys(arr, n, k))`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;``class` `GFG``{``    ` `    ``// Function to return the count of``    ``// maximum toys that can be bought``    ``static` `int` `maxToys(``int``[] arr, ``int` `n, ``int` `k)``    ``{``     ` `        ``// Create a priority_queue and push``        ``// all the array elements in it``        ``List<``int``> pq = ``new` `List<``int``>();``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``pq.Add(arr[i]);``        ``}``        ` `        ``pq.Sort();``     ` `        ``// To store the count of maximum``        ``// toys that can be bought``        ``int` `count = 0;``        ``while` `(pq <= k)``        ``{``            ``count++;``            ``k = k - pq;``            ``pq.RemoveAt(0);``        ``}``        ``return` `count;``    ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] arr = { 1, 12, 5, 111, 200, 1000, 10 };``    ``int` `n = arr.Length;``    ``int` `k = 50;``    ``Console.WriteLine(maxToys(arr, n, k));``  ``}``}` `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N*logN)
Auxiliary Space: O(N)

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