<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]]:
        nums.sort()
        self.res = []
        self.backtrack(nums, [], [False]*len(nums))
        return self.res

    def backtrack(self, nums, perm, used):
        if len(perm) == len(nums):
            self.res.append(perm[:])
            return

        for i in range(len(nums)):
            if used[i]: 
                continue
            if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
                continue
            used[i] = True
            perm.append(nums[i])
            self.backtrack(nums, perm, used)
            perm.pop()
            used[i] = False
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