[Python] Very Very simple python solution
\nclass Solution:\n    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:\n        ## RC ##\n        ## APPROACH : BACKTRACKING ##\n        ## Simila
```\nclass Solution:\n    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:\n        ## RC ##\n        ## APPROACH : BACKTRACKING ##\n        ## Similar to Leetcode : 47. Permutations II ##\n\n\t\t## TIME COMPLEXICITY : O(2^N) ##\n\t\t## SPACE COMPLEXICITY : O(2^N) ##\n\n        def dfs(curr, nums):                          \n            \n            result.append(curr[:])\n\n            if(len(nums) == 0):                                         \n                return\n            \n            for i in range(len(nums)):\n                if i > 0 and nums[i] == nums[i-1]:        # we dont create duplicate initial branches.\n                    continue\n                dfs(curr + [nums[i]], nums[i+1:])         # we are sending, nums[i+1:] for next loop\n                \n        result = []\n        dfs([], sorted(nums))                             \n        return result\n```\nPLEASE UPVOTE IF YOU LIKE MY SOLUTION
['Python', 'Python3']