Sum of average of all subsets

Given an array arr[] of N integer elements, the task is to find the sum of the average of all subsets of this array.

Example:

Input  : arr[] = [2, 3, 5]
Output : 23.33 
Explanation : Subsets with their average are, 
[2]        average = 2/1 = 2
[3]        average = 3/1 = 3
[5]        average = 5/1 = 5
[2, 3]        average = (2+3)/2 = 2.5
[2, 5]        average = (2+5)/2 = 3.5
[3, 5]        average = (3+5)/2 = 4
[2, 3, 5]    average = (2+3+5)/3 = 3.33

Sum of average of all subset is, 
2 + 3 + 5 + 2.5 + 3.5 + 4 + 3.33 = 23.33 

Naive approach: A naive solution is to iterate through all possible subsets, get an average of all of them and then add them one by one, but this will take exponential time and will be infeasible for bigger arrays. 
We can get a pattern by taking an example,  

arr = [a0, a1, a2, a3]
sum of average = 
a0/1 + a1/1 + a2/2 + a3/1 +
(a0+a1)/2 + (a0+a2)/2 + (a0+a3)/2 + (a1+a2)/2 +
 (a1+a3)/2 + (a2+a3)/2 + 
(a0+a1+a2)/3 + (a0+a2+a3)/3 + (a0+a1+a3)/3 + 
 (a1+a2+a3)/3 +
(a0+a1+a2+a3)/4

If S = (a0+a1+a2+a3), then above expression 
can be rearranged as below,
sum of average = (S)/1 + (3*S)/2 + (3*S)/3 + (S)/4

The coefficient with numerators can be explained as follows, suppose we are iterating over subsets with K elements then denominator will be K and the numerator will be r*S, where ‘r’ denotes the number of times a particular array element will be added while iterating over subsets of the same size. By inspection, we can see that r will be nCr(N – 1, n – 1) because after placing one element in summation, we need to choose (n – 1) elements from (N – 1) elements, so each element will have a frequency of nCr(N – 1, n – 1) while considering subsets of the same size, as all elements are taking part in summation equal number of times, this will the frequency of S also and will be the numerator in the final expression. 

In the below code nCr is implemented using dynamic programming method, you can read more about that here, 

C++

// C++ program to get sum of average of all subsets
#include <bits/stdc++.h>
using namespace std;

// Returns value of Binomial Coefficient C(n, k)
int nCr(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;

    // Calculate value of Binomial Coefficient in bottom
    // up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;

            // Calculate value using previously stored
            // values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
    return C[n][k];
}

// method returns sum of average of all subsets
double resultOfAllSubsets(int arr[], int N)
{
    double result = 0.0; // Initialize result

    // Find sum of elements
    int sum = 0;
    for (int i = 0; i < N; i++)
        sum += arr[i];

    // looping once for all subset of same size
    for (int n = 1; n <= N; n++)

        /* each element occurs nCr(N-1, n-1) times while
           considering subset of size n  */
        result += (double)(sum * (nCr(N - 1, n - 1))) / n;

    return result;
}

// Driver code to test above methods
int main()
{
    int arr[] = { 2, 3, 5, 7 };
    int N = sizeof(arr) / sizeof(int);
    cout << resultOfAllSubsets(arr, N) << endl;
    return 0;
}

Java

// java program to get sum of
// average of all subsets
import java.io.*;

class GFG {

    // Returns value of Binomial
    // Coefficient C(n, k)
    static int nCr(int n, int k)
    {
        int C[][] = new int[n + 1][k + 1];
        int i, j;

        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;

                // Calculate value using
                // previously stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
        return C[n][k];
    }

    // method returns sum of average of all subsets
    static double resultOfAllSubsets(int arr[], int N)
    {
        // Initialize result
        double result = 0.0;

        // Find sum of elements
        int sum = 0;
        for (int i = 0; i < N; i++)
            sum += arr[i];

        // looping once for all subset of same size
        for (int n = 1; n <= N; n++)

            /* each element occurs nCr(N-1, n-1) times while
            considering subset of size n */
            result += (double)(sum * (nCr(N - 1, n - 1))) / n;

        return result;
    }

    // Driver code to test above methods
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 5, 7 };
        int N = arr.length;
        System.out.println(resultOfAllSubsets(arr, N));
    }
}

// This code is contributed by vt_m

Python3

# Python3 program to get sum
# of average of all subsets

# Returns value of Binomial
# Coefficient C(n, k)
def nCr(n, k):

    C = [[0 for i in range(k + 1)]
            for j in range(n + 1)]

    # Calculate value of Binomial
    # Coefficient in bottom up manner
    for i in range(n + 1):

        for j in range(min(i, k) + 1):

            # Base Cases
            if (j == 0 or j == i):
                C[i][j] = 1

            # Calculate value using
            # previously stored values
            else:
                C[i][j] = C[i-1][j-1] + C[i-1][j]

    return C[n][k]

# Method returns sum of
# average of all subsets
def resultOfAllSubsets(arr, N):

    result = 0.0 # Initialize result

    # Find sum of elements
    sum = 0
    for i in range(N):
        sum += arr[i]

    # looping once for all subset of same size
    for n in range(1, N + 1):

        # each element occurs nCr(N-1, n-1) times while
        # considering subset of size n */
        result += (sum * (nCr(N - 1, n - 1))) / n

    return result

# Driver code
arr = [2, 3, 5, 7]
N = len(arr)
print(resultOfAllSubsets(arr, N))


# This code is contributed by Anant Agarwal.

C#

// C# program to get sum of
// average of all subsets
using System;

class GFG {

    // Returns value of Binomial
    // Coefficient C(n, k)
    static int nCr(int n, int k)
    {
        int[, ] C = new int[n + 1, k + 1];
        int i, j;

        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.Min(i, k); j++)
            {
                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;

                // Calculate value using
                // previously stored values
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
            }
        }
        return C[n, k];
    }

    // method returns sum of average
    // of all subsets
    static double resultOfAllSubsets(int[] arr, int N)
    {
        // Initialize result
        double result = 0.0;

        // Find sum of elements
        int sum = 0;
        for (int i = 0; i < N; i++)
            sum += arr[i];

        // looping once for all subset
        // of same size
        for (int n = 1; n <= N; n++)

            /* each element occurs nCr(N-1, n-1) times while
               considering subset of size n */
            result += (double)(sum * (nCr(N - 1, n - 1))) / n;

        return result;
    }

    // Driver code to test above methods
    public static void Main()
    {
        int[] arr = { 2, 3, 5, 7 };
        int N = arr.Length;
        Console.WriteLine(resultOfAllSubsets(arr, N));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to get sum
// of average of all subsets

// Returns value of Binomial
// Coefficient C(n, k)
function nCr($n, $k)
{
    $C[$n + 1][$k + 1] = 0;
    $i; $j;

    // Calculate value of Binomial
    // Coefficient in bottom up manner
    for ($i = 0; $i <= $n; $i++)
    {
        for ($j = 0; $j <= min($i, $k); $j++)
        {
            // Base Cases
            if ($j == 0 || $j == $i)
                $C[$i][$j] = 1;

            // Calculate value using
            // previously stored values
            else
                $C[$i][$j] = $C[$i - 1][$j - 1] +
                             $C[$i - 1][$j];
        }
    }
    return $C[$n][$k];
}

// method returns sum of
// average of all subsets
function resultOfAllSubsets($arr, $N)
{
    // Initialize result
    $result = 0.0;

    // Find sum of elements
    $sum = 0;
    for ($i = 0; $i < $N; $i++)
        $sum += $arr[$i];

    // looping once for all
    // subset of same size
    for ($n = 1; $n <= $N; $n++)

        /* each element occurs nCr(N-1,
        n-1) times while considering
        subset of size n */
        $result += (($sum * (nCr($N - 1,
                                 $n - 1))) / $n);

    return $result;
}

// Driver Code
$arr = array( 2, 3, 5, 7 );
$N = sizeof($arr) / sizeof($arr[0]);
echo resultOfAllSubsets($arr, $N) ;

// This code is contributed by nitin mittal.
?>

Javascript

<script>
    // javascript program to get sum of
    // average of all subsets

    // Returns value of Binomial
    // Coefficient C(n, k)
    function nCr(n, k)
    {
        let C = new Array(n + 1);
        for (let i = 0; i <= n; i++)
        {
            C[i] = new Array(k + 1);
            for (let j = 0; j <= k; j++)
            {
                C[i][j] = 0;
            }
        }
        let i, j;

        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;

                // Calculate value using
                // previously stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
        return C[n][k];
    }

    // method returns sum of average of all subsets
    function resultOfAllSubsets(arr, N)
    {
        // Initialize result
        let result = 0.0;

        // Find sum of elements
        let sum = 0;
        for (let i = 0; i < N; i++)
            sum += arr[i];

        // looping once for all subset of same size
        for (let n = 1; n <= N; n++)

            /* each element occurs nCr(N-1, n-1) times while
            considering subset of size n */
            result += (sum * (nCr(N - 1, n - 1))) / n;

        return result;
    }

    let arr = [ 2, 3, 5, 7 ];
    let N = arr.length;
    document.write(resultOfAllSubsets(arr, N));

</script>

Output

63.75

Time Complexity: O(n³)
Auxiliary Space: O(n²)

Efficient Approach : Space Optimization O(1)
To optimize the space complexity of the above approach, we can use a more efficient approach that avoids the need for the entire matrix C[][] to store binomial coefficients. Instead, we can use a combination formula to calculate the binomial coefficient directly when needed.

Implementation steps:

• Iterate over the elements of the array and calculate the sum of all elements.
• Iterate over each subset size from 1 to N.
• Inside the loop, calculate the average of the sum of elements multiplied by the binomial coefficient for the subset size. Add the calculated average to the result.
• Return the final result.

Implementation:

C++

#include <iostream>
using namespace std;

// Method to calculate binomial coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int res = 1;

    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;

    // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1]
    for (int i = 0; i < k; i++)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

// Method to calculate the sum of the average of all subsets
double resultOfAllSubsets(int arr[], int N)
{
    double result = 0.0;
    int sum = 0;

    // Calculate the sum of elements
    for (int i = 0; i < N; i++)
        sum += arr[i];

    // Loop for each subset size
    for (int n = 1; n <= N; n++)
        result += (double)(sum * binomialCoeff(N - 1, n - 1)) / n;

    return result;
}

// Driver code to test the above methods
int main()
{
    int arr[] = { 2, 3, 5, 7 };
    int N = sizeof(arr) / sizeof(int);
    cout << resultOfAllSubsets(arr, N) << endl;
    return 0;
}

Output

63.75

Time Complexity: O(n^2)
Auxiliary Space: O(1)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.