# Maximize the profit by selling at-most M products

Given two lists that contain cost prices CP[] and selling prices SP[] of products respectively. The task is to maximize the profit by selling at-most ‘M’ products.

Examples:

Input: N = 5, M = 3
CP[]= {5, 10, 35, 7, 23}
SP[] = {11, 10, 0, 9, 19}
Output:
Profit on 0th product i.e. 11-5 = 6
Profit on 3rd product i.e. 9-7 = 2
Selling any other product will not give profit.
So, total profit = 6+2 = 8.

Input: N = 4, M = 2
CP[] = {17, 9, 8, 20}
SP[] = {10, 9, 8, 27}
Output:

Approach:

1. Store the profit/loss on buying and selling of each product i.e. SP[i]-CP[i] in an array.
2. Sort that array in descending order.
3. Add the positive values up to M values as positive values denote profit.
4. Return Sum.

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach: #include using namespace std;   // Function to find profit int solve(int N, int M, int cp[], int sp[]) {     int profit[N];       // Calculating profit for each gadget     for (int i = 0; i < N; i++)         profit[i] = sp[i] - cp[i];       // sort the profit array in descending order     sort(profit, profit + N, greater());       // variable to calculate total profit     int sum = 0;       // check for best M profits     for (int i = 0; i < M; i++) {         if (profit[i] > 0)             sum += profit[i];         else             break;     }       return sum; }   // Driver Code int main() {       int N = 5, M = 3;     int CP[] = { 5, 10, 35, 7, 23 };     int SP[] = { 11, 10, 0, 9, 19 };       cout << solve(N, M, CP, SP);       return 0; }

## Java

 // Java implementation of above approach: import java.util.*; import java.lang.*; import java.io.*;   class GFG {   // Function to find profit static int solve(int N, int M,                  int cp[], int sp[]) {     Integer []profit = new Integer[N];       // Calculating profit for each gadget     for (int i = 0; i < N; i++)         profit[i] = sp[i] - cp[i];       // sort the profit array     // in descending order     Arrays.sort(profit, Collections.reverseOrder());       // variable to calculate total profit     int sum = 0;       // check for best M profits     for (int i = 0; i < M; i++)     {         if (profit[i] > 0)             sum += profit[i];         else             break;     }       return sum; }   // Driver Code public static void main(String args[]) {     int N = 5, M = 3;     int CP[] = { 5, 10, 35, 7, 23 };     int SP[] = { 11, 10, 0, 9, 19 };       System.out.println(solve(N, M, CP, SP)); } }   // This code is contributed // by Subhadeep Gupta

## Python3

 # Python3 implementation # of above approach   # Function to find profit def solve(N, M, cp, sp) :           # take empty list     profit = []           # Calculating profit     # for each gadget     for i in range(N) :         profit.append(sp[i] - cp[i])       # sort the profit array     # in descending order     profit.sort(reverse = True)       sum = 0           # check for best M profits     for i in range(M) :         if profit[i] > 0 :             sum += profit[i]         else :             break       return sum   # Driver Code if __name__ == "__main__" :       N, M = 5, 3     CP = [5, 10, 35, 7, 23]     SP = [11, 10, 0, 9, 19]           # function calling     print(solve(N, M, CP, SP))       # This code is contributed # by ANKITRAI1

## C#

 // C# implementation of above approach: using System;   class GFG {   // Function to find profit static int solve(int N, int M,                  int[] cp, int[] sp) {     int[] profit = new int[N];       // Calculating profit for each gadget     for (int i = 0; i < N; i++)         profit[i] = sp[i] - cp[i];       // sort the profit array     // in descending order     Array.Sort(profit);     Array.Reverse(profit);       // variable to calculate total profit     int sum = 0;       // check for best M profits     for (int i = 0; i < M; i++)     {         if (profit[i] > 0)             sum += profit[i];         else             break;     }       return sum; }   // Driver Code public static void Main() {     int N = 5, M = 3;     int[] CP = { 5, 10, 35, 7, 23 };     int[] SP = { 11, 10, 0, 9, 19 };       Console.Write(solve(N, M, CP, SP)); } }   // This code is contributed // by ChitraNayal

## PHP

 0)             \$sum += \$profit[\$i];         else             break;     }       return \$sum; }   // Driver Code \$N = 5; \$M = 3; \$CP = array( 5, 10, 35, 7, 23 ); \$SP = array( 11, 10, 0, 9, 19 );   echo solve(\$N, \$M, \$CP, \$SP);   // This code is contributed // by ChitraNayal ?>

## Javascript



Output

8

Complexity Analysis:

• Time Complexity: O(n*log(n)+m)
• Auxiliary Space: O(n)

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