# Maximize the profit by selling at-most M products in C++

Given the task is to calculate the maximum profit that can be made by selling at-most ‘M’ products.

The total number of products are ‘N’ and the cost price and the selling price of each product is given in the lists CP[] and SP[] respectively.

Input

N=6, M=4
CP[]={1,9,5,8,2,11}
SP[]={1,15,10,16,5,20}

Output

28

Explanation − The profit obtained from selling all the products are 0,6,5,8,3,9 respectively.

So, in order to make maximum profit by selling only 4 products, the products with the highest profit need to be chosen, that is, product number 2,3,4 and 6.

Maximum profit= 6+5+8+9= 28

Input

N=3, M=2
CP[]={10,20,30}
SP[]={19,22,38}

Output

17

## Approach used in the below program as follows

• Create an array Profit[] of type int and size ‘N’ to store the profit obtained from each product.

• Create a variable Total of type int to store the final maximum profit.

• Loop from i=0 till i<N

• While in loop, set Profit[i] = Sp[i] – Cp[i]

• Call function sort(Profit, Profit + N, greater<int>() ); to arrange the Profit[] array in descending array.

• Again loop from i=0 till i<M

• While in loop set an if condition, if(Profit[i]>0) to check if the value positive or not and if so then set total+=Profit[i];

## Example

#include <bits/stdc++.h>
using namespace std;
//Function to find profit
int MaxProfit(int N, int M, int Cp[], int Sp[]){
int Profit[N];
int total = 0;
//Calculating profit from each product
for (int i = 0; i < N; i++)
Profit[i] = Sp[i] - Cp[i];
//Arranging profit array in descending order
sort(Profit, Profit + N, greater<int>());
//Adding the best M number of profits
for (int i = 0; i < M; i++){
if (Profit[i] > 0)
total += Profit[i];
else
break;
}
}
//Main function
int main(){
int MP;
int N=6,M=4;
int CP[] = { 1, 9, 5, 8, 2, 11 };
int SP[] = { 1, 15, 10, 16, 5, 20 };
MP = MaxProfit(N, M, CP, SP);
cout<<”Maximum Profit:”<<MP;
return 0;
}

## Output

If we run the above code we will get the following output −

Maximum Profit: 28