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#90. Subsets II

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	`1`	`+# Given an integer array nums that may contain duplicates, return all possible`
	`2`	`+# subsets`
	`3`	`+# (the power set).`
	`4`	`+`
	`5`	`+# The solution set must not contain duplicate subsets. Return the solution in any order.`
	`6`	`+`
	`7`	`+`
	`8`	`+`
	`9`	`+# Example 1:`
	`10`	`+`
	`11`	`+# Input: nums = [1,2,2]`
	`12`	`+# Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]`
	`13`	`+# Example 2:`
	`14`	`+`
	`15`	`+# Input: nums = [0]`
	`16`	`+# Output: [[],[0]]`
	`17`	`+`
	`18`	`+`
	`19`	`+# Constraints:`
	`20`	`+`
	`21`	`+# 1 <= nums.length <= 10`
	`22`	`+# -10 <= nums[i] <= 10`
	`23`	`+`
	`24`	`+`
	`25`	`+from typing import List`
	`26`	`+class Solution:`
	`27`	`+ def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:`
	`28`	`+ ans = []`
	`29`	`+ def dfs(s, path):`
	`30`	`+ ans.append(path)`
	`31`	`+`
	`32`	`+ for i in range(s, len(nums)):`
	`33`	`+ if i > s and nums[i] == nums[i-1]:`
	`34`	`+ continue`
	`35`	`+ dfs(i+1, path+[nums[i]])`
	`36`	`+ nums.sort()`
	`37`	`+ dfs(0, [])`
	`38`	`+ return ans`

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