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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];
return total;
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; } return total; } //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