<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 make_permutation(self, idx: int, comb: List[int], nums: List[int], dp, ans):
        n = len(nums)
        if (idx, tuple(comb), tuple(nums)) in dp:
            return
        if idx >= n:
            # print("   [Final][COMB]", comb, "[NUMS]", nums)
            ans.add(tuple(comb))
            return
        dp.add((idx, tuple(comb), tuple(nums)))
        # print("[COMB]", comb, "[NUMS]", nums)
        last = None
        for i in range(n):
            # the same number has not replaced the same index before
            if nums[i] != None and nums[i] != last: 
                last = comb[idx] = nums[i]
                nums[i] = None
                self.make_permutation(idx + 1, comb, nums, dp, ans)
                nums[i], comb[idx] = comb[idx], None
        
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        # at each idx, can place one of n numbers
        # each state, a loop till n
        # repeating states? [idx, [nums]] 
        # n * n! * n
        n = len(nums)
        comb = [None] * n
        ans = set()
        self.make_permutation(0, comb, nums, set(), ans)
        return ans
18600