<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>

# O(N * !N) T and O(N * !N) S recursive backtracking solution (NeetCode's modded)
class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        res = []
        N = len(nums)
        numToCount = {}
        for n in nums:
            numToCount[n] = numToCount.get(n, 0) + 1
        
        def dfs(perm):
            if len(perm) >= N:
                res.append(perm[:])
                return
            for n in numToCount.keys():
                if numToCount[n] > 0:
                    perm.append(n)
                    numToCount[n] -= 1
                    dfs(perm)
                    perm.pop()
                    numToCount[n] += 1
        
        dfs([])
        return res
16300