<p>Given a collection of numbers, <code>nums</code>,&nbsp;that might contain duplicates, return <em>all possible unique permutations <strong>in any order</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,2]
<strong>Output:</strong>
[[1,1,2],
 [1,2,1],
 [2,1,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 8</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
</ul>

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        ans = []
        nums.sort()
        dic = set()

        def dfs(index, nums, visited, path, ans):
            print(path)
            visited[index] = True
            curPath = path + [nums[index]]
            if tuple(curPath) in dic:
                visited[index] = False
                return
            dic.add(tuple(curPath))
            if len(curPath) == len(nums):
                ans.append(curPath)
                visited[index] = False
                return

            for i in range(len(nums)):
                print(visited)
                if not visited[i]:
                    dfs(i, nums, visited, curPath, ans)
            visited[index] = False

        for i in range(len(nums)):
            visited = [False] * len(nums)
            dfs(i, nums, visited, [], ans)
        return ans

18000