<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(vector<int> &nums, vector<int> &perm , vector<bool> used , set<vector<int>> &ans, int idx){
        if(idx==nums.size()){
            ans.insert(perm);
            return;
        }
        for(int i=0;i<nums.size();i++){
            if(used[i]==false){
                used[i]=true;
                perm[idx]=nums[i];
                solve(nums , perm , used, ans , idx+1);
                used[i]=false;
            }
        }
    }

    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<int> perm(nums.size());
        vector<bool> used(nums.size(), false);

        set<vector<int>> ans;

        solve(nums,perm,used,ans,0);
        vector<vector<int>> res;
        for(vector<int> v: ans){
            res.push_back(v);
        }
        return res;
    }
};
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