<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 solve(set<vector<int>> &ans, vector<int> nums, int pos) {
        if(pos >= nums.size()-1) {ans.insert(nums); return; }
        for(int i=pos;i<nums.size(); i++) {
            swap(nums[pos], nums[i]);
            int st = pos;
            // while(st<nums.size() && nums[st] == nums[pos]) st++;
            solve(ans, nums, st+1);
            swap(nums[pos], nums[i]);
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        set<vector<int>> ans;
        solve(ans, nums, 0);
        vector<vector<int>> res; res.reserve(ans.size());
        for(auto x: ans) res.push_back(x);
        return res;
    }
};
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