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,  average = 2/1 = 2  average = 3/1 = 3  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 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,
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