<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 __init__(self):
        self.res = []
        self.track = []
        self.visited = []

    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        self.visited = [False] * len(nums)
        self.backtrack(nums)
        return self.res

    def backtrack(self, nums: List[int]) -> None:
        # Ending condition: reaches bottom of tree
        if len(self.track) == len(nums):
            # list() is for empty list situation,
            # Which can be included as list rather None.
            self.res.append(self.track.copy())
            return

        # Go thro all choices:
        for i in range(len(nums)):
            # Prevent repeated element for Permutation
            if self.visited[i]:
                continue
            # Trim the unneccessary repeated elements from the tree
            # Make sure we are comparing at the same level
            if i > 0 and nums[i - 1] == nums[i] and not self.visited[i - 1]:
                continue

            # Preorder: Add element into track
            self.track.append(nums[i])
            self.visited[i] = True
            # Next level
            self.backtrack(nums)
            # Postorder: Discard element from track
            self.track.pop()
            self.visited[i] = False

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