<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 {
public:
    void recursion(set<vector<int>>& v, vector<int>& r, vector<int>& nums, int size)
    {
        if(r.size()==size)
        {
            v.insert(r);
            return;
        }
        if(nums.size()<=0) return;
        for(int i = 0;i<nums.size();i++)
        {
            vector<int> cp = nums;
            r.push_back(nums[i]);
            cp.erase(cp.begin()+i);
            recursion(v, r, cp, size);
            r.pop_back();
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        set<vector<int>> v;
        vector<int> r;
        recursion(v, r, nums, nums.size());
        vector<vector<int>> ans;
        for(auto &i:v) ans.push_back(i);
        return ans;
    }
};
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