<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 get_permutes(self,nums,n,start,elem,ln):
        if not nums or n==0:
            return [[elem]]
        permutes=[]
        for i in range(n):
            # if i>0 and nums[i]==nums[i-1] and i<n-1:
            #     continue
            res=self.get_permutes(nums[0:i]+nums[i+1:],n-1,i+1,nums[i],ln)
            for r in res:
                permutes.append([elem]+r)
        return permutes


    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        d={}
        n=len(nums)
        ans=[]
        for i in range(n):
            # if nums[i] not in d:
            res=self.get_permutes(nums[0:i]+nums[i+1:],n-1,0,nums[i],n)
            for r in res:
                if len(r)==n and r not in ans:
                    ans.append(r)
            d[nums[i]]=1
        
        # print('ans=',ans)
        return ans



        
18100