<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 solve(vector<int>& nums,unordered_map<int,int> freq,vector<vector<int>>& ans,vector<int> temp)
    {
        if(temp.size()==nums.size()) 
        {ans.push_back(temp); return;}

        for(auto i: freq)
        {
            int a=i.first;
            int b=i.second;
            
            if(b>0) 
            {
               freq[a]--;
               temp.push_back(a);
               solve(nums,freq,ans,temp);
               temp.pop_back();
               freq[a]++;
               
            }
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        unordered_map<int,int> freq;
        for(auto a:nums)
        {
            freq[a]++;
        }
        vector<vector<int>> ans;
        solve(nums,freq,ans,{});
        return ans;
    }
};
19006