Maximize sum of means of two subsets formed by dividing given Array into them

Given an array arr[] of size N, the task is to find the maximum sum of means of 2 non-empty subsets of the given array such that each element is part of one of the subsets.

Examples:

Input:  N = 2, arr[] = {1, 3}
Output:  4.00
Explanation: Since there are only two elements, make two subsets as {1} and {3}.
Their sizes are 1 so mean for both will be 1.00 and 3.00 which sums to 4.00

Input:  N = 5, arr[] = {1, 2, 3, 4, 5}
Output: 7.50
Explanation: Since there are 5 elements in the array, divide the array as {1, 2, 3, 4} and {5}. 
Their sizes are 4 and 1 respectively. The mean for the first subset will be (1+2+3+4)/4 = 2.50 
and the mean of the other subset will be 5. Hence the sum of means will be 2.50+5.00 = 7.50
For any other subset division the sum of means will not be more than 7.50. 
Examples: {1, 2, 3} and {4, 5} the means will be (1+2+3)/3 and (4+5)/2 
which is 2.00 and 4.50 respectively which sums to 6.50 which is less than 7.50

Approach: The task can be solved using some observations It can be observed that the maximum sum of means of 2 non-empty subsets can be achieved if one of the subsets contains only the maximum element & the other subset contains the rest of the elements.

Below is the implementation of the above approach: 

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

// Function to return the maximum sum of the
// mean of the sum of two subsets of an array
double maxMeanSubsetSum(int arr[], int N)
{
    // Sorting the array
    sort(arr, arr + N);

    // Stores the sum of array
    double sum = 0;

    // Loop through the array and store
    // sum of elements except the largest
    for (int i = 0; i < N - 1; i++) {
        sum += arr[i];
    }

    // Calculating the mean
    sum = sum / (N - 1);

    // Adding the largest number
    // to the calculated mean
    sum += arr[N - 1];
    return sum;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = 5;

    // Giving output correct to
    // two decimal places
    cout << setprecision(2) << fixed
         << maxMeanSubsetSum(arr, N);
    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to return the maximum sum of the
  // mean of the sum of two subsets of an array
  static double maxMeanSubsetSum(int arr[], int N)
  {
    
    // Sorting the array
    Arrays.sort(arr);

    // Stores the sum of array
    double sum = 0;

    // Loop through the array and store
    // sum of elements except the largest
    for (int i = 0; i < N - 1; i++) {
      sum += arr[i];
    }

    // Calculating the mean
    sum = sum / (N - 1);

    // Adding the largest number
    // to the calculated mean
    sum += arr[N - 1];
    return sum;
  }

  // Driver code
  public static void main (String[] args) {
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = 5;

    // Giving output correct to
    // two decimal places
    System.out.print(String.format("%.2f", maxMeanSubsetSum(arr, N)));
  }
}

// This code is contributed by hrithikgarg03188.

# Python code for the above approach

# Function to return the maximum sum of the
# mean of the sum of two subsets of an array
def maxMeanSubsetSum(arr, N):

    # Sorting the array
    arr.sort()

    # Stores the sum of array
    sum = 0

    # Loop through the array and store
    # sum of elements except the largest
    for i in range(N - 1):
        sum += arr[i]

    # Calculating the mean
    sum = sum / (N - 1)

    # Adding the largest number
    # to the calculated mean
    sum = sum + arr[N - 1]
    return sum

# Driver code
arr = [1, 2, 3, 4, 5]
N = 5

# Giving output correct to
# two decimal places
print("{0:.2f}".format(maxMeanSubsetSum(arr, N)))

# This code is contributed by Potta Lokesh

// C# program for the above approach
using System;

public class GFG
{

  // Function to return the maximum sum of the
  // mean of the sum of two subsets of an array
  static double maxMeanSubsetSum(int[] arr, int N)
  {

    // Sorting the array
    Array.Sort(arr);

    // Stores the sum of array
    double sum = 0;

    // Loop through the array and store
    // sum of elements except the largest
    for (int i = 0; i < N - 1; i++) {
      sum += arr[i];
    }

    // Calculating the mean
    sum = sum / (N - 1);

    // Adding the largest number
    // to the calculated mean
    sum += arr[N - 1];
    return sum;
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int[] arr = { 1, 2, 3, 4, 5 };
    int N = 5;

    // Giving output correct to
    // two decimal places
    Console.Write(string.Format("{0:0.00}", maxMeanSubsetSum(arr, N)));
  }
}

// This code is contributed by code_hunt.

    <script>
        // JavaScript program for the above approach

        // Function to return the maximum sum of the
        // mean of the sum of two subsets of an array
        const maxMeanSubsetSum = (arr, N) => {
        
            // Sorting the array
            arr.sort();

            // Stores the sum of array
            let sum = 0;

            // Loop through the array and store
            // sum of elements except the largest
            for (let i = 0; i < N - 1; i++) {
                sum += arr[i];
            }

            // Calculating the mean
            sum = sum / (N - 1);

            // Adding the largest number
            // to the calculated mean
            sum += arr[N - 1];
            return sum;
        }

        // Driver code
        let arr = [1, 2, 3, 4, 5];
        let N = 5;

        // Giving output correct to
        // two decimal places
        document.write(parseFloat(maxMeanSubsetSum(arr, N)).toFixed(2));

    // This code is contributed by rakeshsahni

    </script>

7.50

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