<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 backtrack(set<vector<int>>& mySet, vector<int>& remaining, vector<int>& curr){

        if(remaining.empty()){

            mySet.insert(curr);
            return;
        }

        for(int i = 0; i < remaining.size(); i++){

            curr.push_back(remaining[i]);
            vector<int> newRemaining = remaining;
            newRemaining.erase(newRemaining.begin() + i);
            backtrack(mySet,newRemaining, curr);
            curr.pop_back();
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> result;
        set<vector<int>> mySet;
        vector<int> curr;
        backtrack(mySet,nums,curr);
        for (const auto& vec : mySet) {
            result.push_back(vec);
        }
        return result;
    }
};
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