<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>& nums,int indx,vector<vector<int>>&res){
       if(indx>=nums.size()){
           res.push_back(nums);
           return ;
       }
       for(int i=indx;i<nums.size();i++){
           swap(nums[indx],nums[i]);
           f(nums,indx+1,res);
           swap(nums[indx],nums[i]);
       }
   }
   
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>>ans,res;
        int indx=0;
        map<vector<int>,int>mp;
        f(nums,indx,res);
        for(int i=0;i<res.size();i++){
            mp[res[i]]++;
        }
        for(auto it:mp){
            ans.push_back(it.first);
        }
        return ans;
    }
};
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