<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:
int n;
unordered_map<int,int>mpp;

    void Backtrack(vector<vector<int>>&res,vector<int>&curr,vector<int>& nums,unordered_map<int,int>cnt){
        if(curr.size()==n){
            res.push_back(curr);
            return ;
        }

        unordered_set<int>st;
      for(int i=0;i<n;i++){
        if(st.find(nums[i])!=st.end()){
            continue;
        }
        if(mpp[nums[i]]== cnt[nums[i]]){
            continue;
        }

        curr.push_back(nums[i]);
        mpp[nums[i]]++;
        st.insert(nums[i]);
        Backtrack(res,curr,nums,cnt);
        curr.pop_back();
        mpp[nums[i]]--;
      }

    }                               
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        n = nums.size();
        vector<vector<int>>res;
        vector<int>curr;
        unordered_map<int,int>cnt;
        for(int i=0;i<n;i++){
            cnt[nums[i]]++;

        }

        Backtrack(res,curr,nums,cnt);
        return res;
    }
};
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