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 ith book. Profit of buying ith 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> |
49
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
Contact Us