Sum of average of all subsets

Given an array arr of N integer elements, the task is to find sum of 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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A naive solution is to iterate through all possible subsets, get 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 numerator will be r*S, where ‘r’ denotes number of times a particular array element will be added while iterating over subsets of 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 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 final expression.
In 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.  
?> 

Output :

63.75

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Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

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C#

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