Find the Maximum sum of the Array by performing the given operations

Given an Array A[] of size N with repeated elements and all array elements are positive, the task is to find the maximum sum by applying the given operations:

• Select any 2 indexes and select 2 integers(say x and y) such that the product of the elements(x and y) is equal to the product of the elements at the chosen index.
• At last, replace the array element with x and y

Examples:

Input: N = 3, A[] = {2, 3, 2}
Output: 14
Explanation:

• i = 1,  j = 2, x = 6,  y = 1 After applying the following operation we get sum = 9(6 + 1 + 2) and array now become {6, 1, 2} now we again chose i and j and will try to find i and j
• i = 1, j = 3 array become {12, 1, 1} sum become 14, thus the maximum sum will remain 14 and we cannot increase further.

Input: N = 2, A = {1, 3}
Output: 4
Explanation: 

• i = 1, j = 2, x = 3, y = 1 After applying the following operation we get sum = 4 array now become {3, 1}. If we again apply the operation then the array will remain the same so the maximum sum comes out to be 4. 

Approach: Implement the idea below to solve the problem:

Notice the operations then we can observe that we are multiplying 2 array elements so at each time we will select one integer as product and other is 1. why? Because this will satisfy our condition for the operation that is given in the question that is product of x and y should be equal to the product of array element at the index we have Chosen. So, we will follow this operation n -1 times and after following the operation n -1 times we will get our final array which consist of n – 1 ones and one element which is equal to product of all the array elements.

Follow the below steps to implement the above idea:

• Calculate the product of the array.
• Add n – 1 to the final product of the array.
• The resulting variable will be the maximum sum possible.

Below is the implementation for the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate maximum sum
void maximumsum(int A[], int n)
{

    int k = 1;

    // Calculating the product of array
    for (int i = 0; i < n; i++) {
        k *= A[i];
    }

    // Adding n-1 ones
    k += n - 1;

    // Printing the final answer
    cout << k << endl;
}

// Driver Code
int main()
{

    int n = 3;
    int A[] = { 2, 3, 2 };

    // Function call
    maximumsum(A, n);

    return 0;
}

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

// Function to calculate maximum sum
static void maximumsum(int A[], int n)
{

    int k = 1;

    // Calculating the product of array
    for (int i = 0; i < n; i++) {
        k *= A[i];
    }

    // Adding n-1 ones
    k += n - 1;

    // Printing the final answer
    System.out.println(k);
}

// Driver code
public static void main(String[] args)
{
    int n = 3;
    int A[] = { 2, 3, 2 };

    // Function call
    maximumsum(A, n);
}
}

Python3

# Python3 implementation of the above approach
def maximumsum(A, n):
    k = 1

    # Calculating the product of array
    for i in range(n):
        k *= A[i]

    # Adding n-1 ones
    k += n - 1

    # Printing the final answer
    print(k)

# Driver Code
if __name__ == '__main__':
    n = 3
    A = [2, 3, 2]

    # Function call
    maximumsum(A, n)

    # This code is contributed by phasing17.

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;

class Gfg
{
  // Function to calculate maximum sum
  static void maximumsum(int[] A, int n)
  {

    int k = 1;

    // Calculating the product of array
    for (int i = 0; i < n; i++) {
      k *= A[i];
    }

    // Adding n-1 ones
    k += n - 1;

    // Printing the final answer
    Console.WriteLine(k);
  }

  // Driver Code
  public static void Main(string[] args)
  {

    int n = 3;
    int[] A = { 2, 3, 2 };

    // Function call
    maximumsum(A, n);

  }
}

// This code is contributed by imruhrbf8.

Javascript

// Javascript implementation of the above approach

// Function to calculate maximum sum
function maximumsum(A, n)
{

    let k = 1;

    // Calculating the product of array
    for (let i = 0; i < n; i++) {
        k *= A[i];
    }

    // Adding n-1 ones
    k += n - 1;

    // Printing the final answer
    console.log(k);
}

// Driver Code
let n = 3;
let A = [ 2, 3, 2 ];

// Function call
maximumsum(A, n);

Output

14

Time Complexity: O(N) 
Auxiliary Space: O(1)