<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:
    vector<vector<int>> ans;
    unordered_map<int,int> freq;
    void permute(vector<int> res,int x) {
            if(x==res.size()) {
                ans.push_back(res);
                return ;
            }
            for(auto& i:freq) {
                if(i.second>0){
                res.push_back(i.first);
                i.second--;
                permute(res,x);
                res.pop_back();
                i.second++;
                }
            }
        }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        ios_base::sync_with_stdio(false);
        for(int i=0;i<nums.size();i++) {
            freq[nums[i]]++;
        }
        permute({},nums.size());
        return ans;
    }
};
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