Permutations II
<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>
Array; Backtracking; Sorting
class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        def dfs(i: int):
            if i == n:
                ans.append(t[:])
                return
            for j in range(n):
                if vis[j] or (j and nums[j] == nums[j - 1] and not vis[j - 1]):
                    continue
                t[i] = nums[j]
                vis[j] = True
                dfs(i + 1)
                vis[j] = False

        n = len(nums)
        nums.sort()
        ans = []
        t = [0] * n
        vis = [False] * n
        dfs(0)
        return ans
