<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 Permu(vector<int>& nums,int i,int n,vector<vector<int>>&res,unordered_map<string,int>&m)
{
    //base
    if(i==n)
    {
        string str="";
        //convert permutation into string 
        for(auto ele:nums)
        {
            //here works as delimiter for element of nums
            str+=to_string(ele)+",";
        }

        if(m.find(str)==m.end()) //equal to end means not present
        {
            res.push_back(nums);
            m[str]=1;  //make that permu as visited
        }
        else
        {
            //skip this permu as it is already present
            m[str]++;
        }
        return;
    }

    for(int j=i;j<n;j++)
    {
        swap(nums[i],nums[j]);
        Permu(nums,i+1,n,res,m);
        swap(nums[i],nums[j]);
    }
}

vector<vector<int>> permuteUnique(vector<int>& nums) {
        int n=nums.size();
        vector<vector<int>>res;
        unordered_map<string,int>m;
        Permu(nums,0,n,res,m);
        return res;
 }

};
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