<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 swap(int& x, int& y) 
{ 
    int temp = x; 
    x = y; 
    y = temp; 
} 

set< vector<int> >s;

// Function to find the possible 
// permutations 
void permutations(vector<vector<int> >& res, 
                  vector<int> nums, int l, int h) 
{ 
    // Base case 
    // Add the vector to result and return 
    if (l == h) { 
        if(s.find(nums)==s.end())
        {
            res.push_back(nums); 
            s.insert(nums);
        }
        return; 
    } 
  
    // Permutations made 
    for (int i = l; i <= h; i++) { 
  
        // Swapping 
        swap(nums[l], nums[i]); 
  
        // Calling permutations for 
        // next greater value of l 
        permutations(res, nums, l + 1, h); 
  
        // Backtracking 
        swap(nums[l], nums[i]); 
    } 
} 
  
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector< vector<int> >result;
        sort(nums.begin(),nums.end());
        permutations(result,nums,0,nums.size()-1);
        return result;
    }
};
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