the code recursively calculate the subset. The all subsets of the list without the first elements once calculated as L1, the answer would be L1+S[0]+all list in L1. That's the answer for the case without duplicates. When duplicates exists, the duplicated subset need to be removed. To avoid the redundant calculation, the code filters out the duplicates only when they exist. That is when S[0]==S[1](EG. S=[1,1,2]).

```
class Solution:
# @param num, a list of integer
# @return a list of lists of integer
def subsetsWithDup(self, S):
res=[[]]
if S:
S.sort()
subSet=self.subsetsWithDup(S[1:])
if len(S)>1 and S[0]==S[1]:
# duplicates exists only when first two elements are identical
res=subSet+[[S[0]]+set for set in subSet if (([S[0]]+set) not in subSet)]
else:
res=subSet+[[S[0]]+set for set in subSet]
return res
```