<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 genPerms(vector<int> A, vector<int> visited, set<vector<int>> &ans, int n, int count, vector<int> temp){
    void genPerms(vector<int> A, vector<int> visited, vector<vector<int>> &ans, int n, int count, vector<int> temp){
        if(count==n){
            // ans.push_back(temp);
            ans.emplace_back(temp);
            // cout << temp << endl;
            return;
        }
        for(int i=0; i<n; i++){
            // BOTH, visited !visited work, 1 , ! is fast
            // if(i>0 && A[i-1]==A[i] && !visited[i-1]) continue;
            if(i>0 && A[i-1]==A[i] && visited[i-1]) continue;
            // if(i>0 && A[i-1]==A[i]) continue;
            if(visited[i]==0){
                visited[i]=1;
                // temp.push_back(A[i]);
                temp.emplace_back(A[i]);
                genPerms(A, visited, ans, n, count+1, temp);
                temp.pop_back();
                visited[i]=0;
            }
        }
    }
    
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> visited(nums.size(), 0);
        sort(nums.begin(), nums.end());
        vector<int> temp;
        genPerms(nums, visited, ans, nums.size(), 0, temp);
        return ans;
    }
};
22253