Maximum profit that can be obtained by buying at most K books

Given an integer K and an array arr[] consisting of N integers, where an array element arr[i] represents the price of the i^th book. Profit of buying i^th book represents max(0, -1 * arr[i]), the task is to find the maximum profit possible by buying at most K books.

Examples:

Input: arr[] = {-10, 20, -30, 50, -19}, K = 2
Output: 49
Explanation: 
Maximum profit can be obtained by buying the books arr[2](= -30). Profit = 30 and the book, arr[4](= -19) for the profit of 19.
Therefore, the total maximum profit obtained is, (30+19 = 49).

Input: arr[] = {10, 20, 16, 25}, K = 3
Output: 0

Approach: The problem can be solved using the greedy approach based on the observation that only books with negative prices contribute to the maximum profit. Follow the steps below to solve this problem:

• Sort the array arr[] in ascending order.
• Initialize a variable, say, maxBenefit as 0 to store the maximum profit.
• Iterate in the range [0, N-1] using the variable i and perform the following steps:
  • If K is greater than 0 and arr[i] is negative, then add the abs(arr[i]) to maxBenefit and then decrement the value of K by 1.
• Finally, print the value of maxBenefit as the maximum profit obtained.

Below is the implementation of the above approach:

C++

// C++ program for above approach

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

// Function to find the maximum
// profit that can be obtained
// by buying at most K books
int maxProfit(int arr[], int N, int K)
{

    // Sort the array in
    // ascending order
    sort(arr, arr + N);

    // Stores the maximum profit
    int maxBenefit = 0;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {

        // If arr[i] is less than 0
        // and K is greater than 0
        if (arr[i] < 0 && K > 0) {

            // Increment the maxBenefit
            // by abs(arr[i])
            maxBenefit += abs(arr[i]);

            // Decrement K by 1
            K--;
        }
    }

    // Return the profit
    return maxBenefit;
}

// Driver Code
int main()
{

    // Given Input
    int arr[] = { -10, 20, -30, 50, -19 };
    int K = 2;
    int N = sizeof(arr) / sizeof(int);

    // Function call
    cout << maxProfit(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.Arrays;

class GFG
{

  // Function to find the maximum
// profit that can be obtained
// by buying at most K books
    public static int maxProfit(int arr[], int N, int K)
    {

        // Sort the array in
        // ascending order
        Arrays.sort(arr);

        // Stores the maximum profit
        int maxBenefit = 0;

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {

            // If arr[i] is less than 0
            // and K is greater than 0
            if (arr[i] < 0 && K > 0) {

                // Increment the maxBenefit
                // by abs(arr[i])
                maxBenefit += Math.abs(arr[i]);

                // Decrement K by 1
                K--;
            }
        }

        // Return the profit
        return maxBenefit;
    }

// Driver Code
    public static void main(String[] args)
    {

      // Given input
        int arr[] = { -10, 20, -30, 50, -19 };
        int K = 2;
        int N = 5;

        // Function call
        int res = maxProfit(arr, N, K);
        System.out.println(res);
    }
}

// This code is contributed by lokeshpotta20.

Python3

# Python3 program for above approach

# Function to find the maximum
# profit that can be obtained
# by buying at most K books
def maxProfit(arr, N, K):

    # Sort the array in
    # ascending order
    arr.sort()

    # Stores the maximum profit
    maxBenefit = 0

    # Traverse the array arr[]
    for i in range(0, N, 1):

        # If arr[i] is less than 0
        # and K is greater than 0
        if (arr[i] < 0 and K > 0):

            # Increment the maxBenefit
            # by abs(arr[i])
            maxBenefit += abs(arr[i])

            # Decrement K by 1
            K -= 1

    # Return the profit
    return maxBenefit

# Driver Code
if __name__ == '__main__':

    # Given Input
    arr = [ -10, 20, -30, 50, -19 ]
    K = 2
    N = len(arr)

    # Function call
    print(maxProfit(arr, N, K))

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;

class GFG {

    // Function to find the maximum
    // profit that can be obtained
    // by buying at most K books
    public static int maxProfit(int[] arr, int N, int K)
    {

        // Sort the array in
        // ascending order
        Array.Sort(arr);

        // Stores the maximum profit
        int maxBenefit = 0;

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {

            // If arr[i] is less than 0
            // and K is greater than 0
            if (arr[i] < 0 && K > 0) {

                // Increment the maxBenefit
                // by abs(arr[i])
                maxBenefit += Math.Abs(arr[i]);

                // Decrement K by 1
                K--;
            }
        }

        // Return the profit
        return maxBenefit;
    }

    // Driver Code
    public static void Main()
    {

        // Given input
        int[] arr = { -10, 20, -30, 50, -19 };
        int K = 2;
        int N = 5;

        // Function call
        int res = maxProfit(arr, N, K);
        Console.Write(res);
    }
}

// This code is contributed by subhammahato348.

Javascript

<script>
        // JavaScript program for above approach

        // Function to find the maximum
        // profit that can be obtained
        // by buying at most K books
        function maxProfit(arr, N, K) {

            // Sort the array in
            // ascending order
            arr.sort(function (a, b) { return a - b });
            // Stores the maximum profit
            var maxBenefit = 0;

            // Traverse the array arr[]
            for (let i = 0; i < N; i++) {

                // If arr[i] is less than 0
                // and K is greater than 0
                if (arr[i] < 0 && K > 0) {

                    // Increment the maxBenefit
                    // by abs(arr[i])
                    maxBenefit += Math.abs(arr[i]);

                    // Decrement K by 1
                    K--;
                }
            }

            // Return the profit
            return maxBenefit;
        }

        // Driver Code

        // Given Input
        var arr = [-10, 20, -30, 50, -19];
        var K = 2;
        var N = 5;

        // Function call
        document.write(maxProfit(arr, N, K));

// This code is contributed by lokeshpotta20.

    </script>

Output:

49

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)