<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 f(vector<int> &v, vector<int> &curr, set<vector<int>> &sol, int n){
        if(curr.size() == n){
            sol.insert(curr);
            return;
        }
        for(int i = 0; i < v.size(); i++){
            curr.push_back(v[i]);
            vector<int> rem;
            for(int j = 0; j < i; j++){
                rem.push_back(v[j]);
            }
            for(int j = i+1; j < v.size(); j++){
                rem.push_back(v[j]);
            }
            f(rem, curr, sol,n);
            curr.pop_back();
        }
        
    }

    vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<int> curr;
        set<vector<int>> sol;
        f(nums, curr, sol, nums.size());
        
        vector<vector<int>> ans;
        for(auto i: sol) ans.push_back(i);
        return ans;
    }
};
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