<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>> res;
    vector<bool> vis;
    void f(vector<int> &arr , vector<int> &cur){
        if(cur.size() == arr.size()){
            res.push_back(cur); 
            return;
        }else{
            map<int,int> called;
            for(int i=0;i<arr.size();i++){
                if((!vis[i]) && (!called[arr[i]])){
                    cur.push_back(arr[i]);
                    vis[i] = true;  
                    called[arr[i]] = 1;
                    f(arr , cur);
                    cur.pop_back();
                    vis[i] = false;
                }
            }
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<int> cur;
        vis.resize(nums.size() , false);
        f(nums , cur);
        return res;
    }
};
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