Select numbers in such way to maximize the amount of money

Given two arrays A1 and A2 of N numbers. There are two people A and B who select numbers out of N. If A selects i-th number, then he will be paid A1[i] amount of money and if B selects i-th number then he will be paid A2[i] amount of money but A cannot select more than X numbers and B cannot select more than Y numbers. The task is to select N numbers in such a way that the amount of total money is maximized at the end. 
Note: X + Y >= N 

Examples:
Input: N = 5, X = 3, Y = 3 
       A1[] = {1, 2, 3, 4, 5}, 
       A2= {5, 4, 3, 2, 1}
Output: 21 
B will take the first 3 orders and A 
will take the last two orders. 

Input: N = 2, X = 1, Y = 1 
       A1[] = {10, 10}, A2= {20, 20}
Output: 30 

Approach: Let us create a new array C such that C[i] = A2[i] – A1[i]. Now we will sort the array C in decreasing order. Note that the condition X + Y >= N guarantees that we will be able to assign the number to any one of the persons. Assume that for some i, A1[i] > A2[i] and you assigned an order to B, the loss encountered due to this assignment is C[i]. Similarly, for some i, A2[i] > A1[i] and you assigned a number to A, the loss encountered due to this assignment is C[i]. As we want to minimize the loss encountered, it is better to process the numbers having high possible losses, because we can try to reduce the loss in the starting part. There is no point in selecting a number with a high loss after assigning a number with less loss. Hence we initially assign all numbers to A initially, then subtract the loss from them greedily. Once the assigned order number is under X, then we store the maximal of that. 
Below is the implementation of the above approach: 

C++

// C++ program to maximize profit

#include <bits/stdc++.h>
using namespace std;

// Function that maximizes the sum
int maximize(int A1[], int A2[], int n,
             int x, int y)
{
    // Array to store the loss
    int c[n];

    // Initial sum
    int sum = 0;

    // Generate the array C
    for (int i = 0; i < n; i++) {
        c[i] = A2[i] - A1[i];
        sum += A1[i];
    }

    // Sort the array elements
    // in descending order
    sort(c, c + n, greater<int>());

    // Variable to store the answer
    int maxi = -1;

    // Iterate in the array, C
    for (int i = 0; i < n; i++) {

        // Subtract the loss
        sum += c[i];

        // Check if X orders are going
        // to be used
        if (i + 1 >= (n - x))
            maxi = max(sum, maxi);
    }

    return maxi;
}

// Driver Code
int main()
{
    int A1[] = { 1, 2, 3, 4, 5 };
    int A2[] = { 5, 4, 3, 2, 1 };

    int n = 5;
    int x = 3, y = 3;

    cout << maximize(A1, A2, n, x, y);

    return 0;
}

Java

// Java program to maximize profit
import java.util.*;

class GFG
{

// Function that maximizes the sum
static int maximize(int A1[], int A2[], int n,
            int x, int y)
{
    // Array to store the loss
    int[] c = new int[n];

    // Initial sum
    int sum = 0;

    // Generate the array C
    for (int i = 0; i < n; i++)
    {
        c[i] = A2[i] - A1[i];
        sum += A1[i];
    }

    // Sort the array elements
    // in descending order
int temp;
for(int i = 0; i < n - 1; i++)
{
    if(c[i] < c[i+1])
    {
        temp = c[i];
        c[i] = c[i + 1];
        c[i + 1] = temp;
    }
}

    // Variable to store the answer
    int maxi = -1;

    // Iterate in the array, C
    for (int i = 0; i < n; i++)
    {

        // Subtract the loss
        sum += c[i];

        // Check if X orders are going
        // to be used
        if (i + 1 >= (n - x))
            maxi = Math.max(sum, maxi);
    }

    return maxi;
}

// Driver Code
public static void main(String args[])
{
    int A1[] = { 1, 2, 3, 4, 5 };
    int A2[] = { 5, 4, 3, 2, 1 };

    int n = 5;
    int x = 3, y = 3;

    System.out.println(maximize(A1, A2, n, x, y));
}
}

// This code is contributed by
// Surendra_Gangwar

Python3

# Python3 program to maximize profit

# Function that maximizes the Sum
def maximize(A1, A2, n, x, y):

    # Array to store the loss
    c = [0 for i in range(n)]

    # Initial Sum
    Sum = 0

    # Generate the array C
    for i in range(n):
        c[i] = A2[i] - A1[i]
        Sum += A1[i]

    # Sort the array elements
    # in descending order
    c.sort()
    c = c[::-1]

    # Variable to store the answer
    maxi = -1

    # Iterate in the array, C
    for i in range(n):

        # Subtract the loss
        Sum += c[i]

        # Check if X orders are going
        # to be used
        if (i + 1 >= (n - x)):
            maxi = max(Sum, maxi)

    return maxi

# Driver Code
A1 = [ 1, 2, 3, 4, 5 ]
A2 = [ 5, 4, 3, 2, 1 ]

n = 5
x, y = 3, 3

print(maximize(A1, A2, n, x, y))

# This code is contributed
# by Mohit Kumar

C#

// C# program to maximize profit
using System;

class GFG
{

    // Function that maximizes the sum
    static int maximize(int [] A1, int [] A2, int n,
                                    int x, int y)
    {
        // Array to store the loss
        int [] c = new int[n];

        // Initial sum
        int sum = 0;

        // Generate the array C
        for (int i = 0; i < n; i++)
        {
            c[i] = A2[i] - A1[i];
            sum += A1[i];
        }

        // Sort the array elements
        // in descending order
            int temp;
        for(int i = 0; i < n - 1; i++)
        {
            if(c[i] < c[i+1])
            {
                temp = c[i];
                c[i] = c[i + 1];
                c[i + 1] = temp;
            }
        }

        // Variable to store the answer
        int maxi = -1;

        // Iterate in the array, C
        for (int i = 0; i < n; i++)
        {

            // Subtract the loss
            sum += c[i];

            // Check if X orders are going
            // to be used
            if (i + 1 >= (n - x))
                maxi = Math.Max(sum, maxi);
        }

        return maxi;
    }

    // Driver Code
    public static void Main()
    {
        int [] A1 = { 1, 2, 3, 4, 5 };
        int [] A2 = { 5, 4, 3, 2, 1 };

        int n = 5;
        int x = 3, y = 3;

        Console.WriteLine(maximize(A1, A2, n, x, y));
    }
}

// This code is contributed by ihritik

PHP

<?php
// PHP program to maximize profit

// Function that maximizes the sum
function maximize($A1, $A2, $n, $x, $y)
{
    # Array to store the loss
    $c = array();

    # Initial sum
    $sum = 0;

    // Generate the array C
    for ($i = 0; $i < $n; $i++)
    {
        $c[$i] = $A2[$i] - $A1[$i];
        $sum += $A1[$i];
    }

    // Sort the array elements
    // in descending order
    rsort($c);

    // Variable to store the answer
    $maxi = -1;

    // Iterate in the array, C
    for ($i = 0; $i < $n; $i++)
    {

        // Subtract the loss
        $sum += $c[$i];

        // Check if X orders are going
        // to be used
        if ($i + 1 >= ($n - $x))
            $maxi = max($sum, $maxi);
    }

    return $maxi;
}

# Driver Code
$A1 = array( 1, 2, 3, 4, 5 );
$A2 = array( 5, 4, 3, 2, 1 );

$n = 5;
$x = 3;
$y = 3;

echo maximize($A1, $A2, $n, $x, $y);

// This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation of the approach

// Function that maximizes the sum
function maximize(A1, A2, n,
            x, y)
{
    // Array to store the loss
    let c = Array(n).fill(0);

    // Initial sum
    let sum = 0;

    // Generate the array C
    for (let i = 0; i < n; i++)
    {
        c[i] = A2[i] - A1[i];
        sum += A1[i];
    }

    // Sort the array elements
    // in descending order
let temp;
for(let i = 0; i < n - 1; i++)
{
    if(c[i] < c[i+1])
    {
        temp = c[i];
        c[i] = c[i + 1];
        c[i + 1] = temp;
    }
}

    // Variable to store the answer
    let maxi = -1;

    // Iterate in the array, C
    for (let i = 0; i < n; i++)
    {

        // Subtract the loss
        sum += c[i];

        // Check if X orders are going
        // to be used
        if (i + 1 >= (n - x))
            maxi = Math.max(sum, maxi);
    }

    return maxi;
}


// Driver code

     let A1 = [ 1, 2, 3, 4, 5 ];
     let A2 = [ 5, 4, 3, 2, 1 ];

    let n = 5;
    let x = 3, y = 3;

    document.write(maximize(A1, A2, n, x, y));

</script>

Output:

21

Time Complexity: O(N*logN), as we are using an inbuilt sort function that costs N*logN time.

Auxiliary Space: O(N), as we are using extra space of the array c. Where N is the number of elements in the arrays.