Maximum items that can be bought from the cost Array based on given conditions

Given an array arr[] of size N where every index in the array represents the cost of buying an item and two numbers P, K. The task is to find the maximum number of items which can be bought such that: 

1. If some i-th object is bought from the array, the remaining amount becomes P – arr[i].
2. We can buy K items, not necessarily consecutive, at a time by paying only for the item whose cost is maximum among them. Now, the remaining amount would be P – max(cost of K items).

Examples:

Input: arr[] = {2, 4, 3, 5, 7}, P = 6, K = 2
Output: 3 
Explanation:
We can buy the first item whose cost is 2. So, the remaining amount is P = 6 – 2 = 4. 
Now, we can choose the second and third item and pay for the maximum one which is max(4, 3) = 4, and the remaining amount is 4 – 4 = 0. 
Therefore, the total number of items bought is 3. 

Input: arr[] = {2, 4, 3, 5, 7}, P = 11, K = 2
Output: 4 
Explanation:
We can buy the first and third item together and pay for only the maximum one which is max(2, 3) = 3. The remaining amount is P = 11 – 3 = 8. 
Now, we can buy the second and fourth item and pay for the maximum one which is max(4, 5) = 5. The remaining amount is P = 8 – 5 = 3. Now, we cant buy any item further. 

Approach: The idea is to use the concept of sorting and prefix sum array. 

• Sort the given array arr[].
• Find the prefix sum for the array arr[].
• The idea behind sorting is that the maximum number of items can be bought only when we buy the items with less cost. This type of algorithm is known as a greedy algorithm.
• And, we use the prefix sum array to find the cost of buying the items.

Below is the implementation of the above approach: 

C++

// C++ program to find the
// maximum number of items
// that can be bought from
// the given cost array

#include <bits/stdc++.h>
using namespace std;

// Function to find the
// maximum number of items
// that can be bought from
// the given cost array
int number(int a[], int n, int p, int k)
{
    // Sort the given array
    sort(a, a + n);

    // Variables to store the prefix
    // sum, answer and the counter
    // variables
    int pre[n] = { 0 }, val, i,
        j, ans = 0;

    // Initializing the first element
    // of the prefix array
    pre[0] = a[0];

    // If we can buy at least one item
    if (pre[0] <= p)
        ans = 1;

    // Iterating through the first
    // K items and finding the
    // prefix sum
    for (i = 1; i < k - 1; i++) {
        pre[i] = pre[i - 1] + a[i];

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }

    pre[k - 1] = a[k - 1];

    // Finding the prefix sum for
    // the remaining elements
    for (i = k - 1; i < n; i++) {
        if (i >= k) {
            pre[i] += pre[i - k] + a[i];
        }

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }

    return ans;
}

// Driver code
int main()
{
    int n = 5;
    int arr[] = { 2, 4, 3, 5, 7 };
    int p = 11;
    int k = 2;

    cout << number(arr, n, p, k) << endl;

    return 0;
}

Java

// Java program to find the maximum
// number of items that can be bought
// from the given cost array
import java.io.*;
import java.util.*;

class GFG{

// Function to find the
// maximum number of items
// that can be bought from
// the given cost array
static int number(int[] a, int n,
                  int p, int k)
{

    // Sort the given array
    Arrays.sort(a);

    // Variables to store the prefix
    // sum, answer and the counter
    // variables
    int[] pre = new int[n];
    int val, i, j, ans = 0;

    // Initializing the first element
    // of the prefix array
    pre[0] = a[0];

    // If we can buy at least one item
    if (pre[0] <= p)
        ans = 1;

    // Iterating through the first
    // K items and finding the
    // prefix sum
    for(i = 1; i < k - 1; i++)
    {
        pre[i] = pre[i - 1] + a[i];

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }
    pre[k - 1] = a[k - 1];

    // Finding the prefix sum for
    // the remaining elements
    for(i = k - 1; i < n; i++)
    {
        if (i >= k)
        {
            pre[i] += pre[i - k] + a[i];
        }

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }
    return ans;
}

// Driver code
public static void main(String[] args)
{
    int n = 5;
    int[] arr = { 2, 4, 3, 5, 7 };
    int p = 11;
    int k = 2;

    System.out.println(number(arr, n, p, k));
}
}

// This code is contributed by akhilsaini

Python3

# Python3 program to find the maximum
# number of items that can be bought
# from the given cost array

# Function to find the maximum
# number of items that can be
# bought from the given cost array
def number(a, n, p, k):

    # Sort the given array
    a.sort()

    # Variables to store the prefix
    # sum, answer and the counter
    # variables
    pre = [ ]
    for i in range(n):
        pre.append(0)

    ans = 0
    val = 0
    i = 0
    j = 0

    # Initializing the first element
    # of the prefix array
    pre[0] = a[0]

    # If we can buy at least one item
    if pre[0] <= p:
        ans = 1

    # Iterating through the first
    # K items and finding the
    # prefix sum
    for i in range(1, k - 1):
        pre[i] = pre[i - 1] + a[i]

        # Check the number of items
        # that can be bought
        if pre[i] <= p:
            ans = i + 1

    pre[k - 1] = a[k - 1]

    # Finding the prefix sum for
    # the remaining elements
    for i in range(k - 1, n):
        if i >= k:
            pre[i] += pre[i - k] + a[i]

        # Check the number of items
        # that can be bought
        if pre[i] <= p:
            ans = i+ 1

    return ans

# Driver code
n = 5
arr = [ 2, 4, 3, 5, 7 ]
p = 11
k = 2

print(number(arr, n, p, k))

# This code is contributed by ishayadav181

C#

// C# program to find the maximum
// number of items that can be
// bought from the given cost array
using System;
using System.Collections;

class GFG{

// Function to find the
// maximum number of items
// that can be bought from
// the given cost array
static int number(int[] a, int n,
                  int p, int k)
{

    // Sort the given array
    Array.Sort(a);

    // Variables to store the prefix
    // sum, answer and the counter
    // variables
    int[] pre = new int[n];
    int i, ans = 0;

    // Initializing the first element
    // of the prefix array
    pre[0] = a[0];

    // If we can buy at least one item
    if (pre[0] <= p)
        ans = 1;

    // Iterating through the first
    // K items and finding the
    // prefix sum
    for(i = 1; i < k - 1; i++)
    {
        pre[i] = pre[i - 1] + a[i];

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }

    pre[k - 1] = a[k - 1];

    // Finding the prefix sum for
    // the remaining elements
    for(i = k - 1; i < n; i++)
    {
        if (i >= k)
        {
            pre[i] += pre[i - k] + a[i];
        }

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }
    return ans;
}

// Driver code
static public void Main ()
{
    int n = 5;
    int[] arr = { 2, 4, 3, 5, 7 };
    int p = 11;
    int k = 2;

    Console.WriteLine(number(arr, n, p, k));
}
}

// This code is contributed by akhilsaini

Javascript

<script>

// Javascript program to find the
// maximum number of items
// that can be bought from
// the given cost array

// Function to find the
// maximum number of items
// that can be bought from
// the given cost array
function number(a, n, p, k)
{
    // Sort the given array
    a.sort();

    // Variables to store the prefix
    // sum, answer and the counter
    // variables
    var pre = Array(n).fill(0), val, i,
        j, ans = 0;

    // Initializing the first element
    // of the prefix array
    pre[0] = a[0];

    // If we can buy at least one item
    if (pre[0] <= p)
        ans = 1;

    // Iterating through the first
    // K items and finding the
    // prefix sum
    for (i = 1; i < k - 1; i++) {
        pre[i] = pre[i - 1] + a[i];

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }

    pre[k - 1] = a[k - 1];

    // Finding the prefix sum for
    // the remaining elements
    for (i = k - 1; i < n; i++) {
        if (i >= k) {
            pre[i] += pre[i - k] + a[i];
        }

        // Check the number of items
        // that can be bought
        if (pre[i] <= p)
            ans = i + 1;
    }

    return ans;
}

// Driver code
var n = 5;
var arr = [2, 4, 3, 5, 7];
var p = 11;
var k = 2;
document.write( number(arr, n, p, k));

</script>

Output:

4

Time Complexity: O(N*logN)

Auxiliary Space: O(N)