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;
} |
// 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 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# 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. |
<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> |
4
Time Complexity: O(N*logN)
Auxiliary Space: O(N)