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, 

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:

Output

63.75

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