# Maximize the profit by selling at-most M products

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

Examples:

Input: N = 5, M = 3
CP[]= {5, 10, 35, 7, 23}
SP[] = {11, 10, 0, 9, 19}
Output: 8
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: 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 decending order ` `    ``sort(profit, profit + N, greater<``int``>()); ` ` `  `    ``// 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 decending 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 decending 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 ` `?> `

Output:

```8
```

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