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

Difficulty Level : Easy
Last Updated : 13 Apr, 2021

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++

// 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;
}

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<Integer> pq = new PriorityQueue<Integer>();
    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
def maxToys(arr, n, k) :
   
    # Create a priority_queue and push
    # all the array elements in it
    pq = []
    for i in range(n) :
        pq.append(arr[i])
    pq.sort()
   
    # To store the count of maximum
    # toys that can be bought
    count = 0
    while (pq[0] <= k) :
        count += 1
        k = k - pq[0]
        pq.pop(0)
    return count
     
    # Driver code
arr = [ 1, 12, 5, 111, 200, 1000, 10 ]
n = len(arr)
k = 50
print(maxToys(arr, n, k))
 
# This code is contributed by divyesh072019

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[0] <= k)
        {
            count++;
            k = k - pq[0];
            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

<script>
 
    // Javascript implementation of the approach
     
    // Function to return the count of
    // maximum toys that can be bought
    function maxToys(arr, n, k)
    {
       
        // Create a priority_queue and push
        // all the array elements in it
        let pq = [];
        for (let i = 0; i < n; i++)
        {
            pq.push(arr[i]);
        }
          
        pq.sort(function(a, b){return a - b});
       
        // To store the count of maximum
        // toys that can be bought
        let count = 0;
        while (pq[0] <= k)
        {
            count++;
            k = k - pq[0];
            pq.shift();
        }
        return count;
    }
       
    let arr = [ 1, 12, 5, 111, 200, 1000, 10 ];
    let n = arr.length;
    let k = 50;
    document.write(maxToys(arr, n, k));
     
</script>

Output:

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