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

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

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 IfS= (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

Output:

63.75

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