<p>Given an integer array <code>nums</code> that may contain duplicates, return <em>all possible</em> <span data-keyword="subset"><em>subsets</em></span><em> (the power set)</em>.</p>

<p>The solution set <strong>must not</strong> contain duplicate subsets. Return the solution in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,2]
<strong>Output:</strong> [[],[1],[1,2],[1,2,2],[2],[2,2]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [0]
<strong>Output:</strong> [[],[0]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
</ul>

class Solution {
public:
void recur(int ind,vector<int>&v,vector<vector<int>> &s,vector<int> &nums,int n){
    if(ind==n){
        s.push_back(v);
            return ;
    }
          v.push_back(nums[ind]);
          recur(ind+1,v,s,nums,n);
          v.pop_back();
          recur(ind+1,v,s,nums,n); 
}
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        int n=nums.size();
        vector<vector<int>>s;
        set<vector<int>>se;
        vector<int>v;
        recur(0,v,s,nums,n);
        vector<vector<int>>ans;
   for(int i=0;i<s.size();i++)  {
      sort(s[i].begin(),s[i].end());
      se.insert(s[i]);
   }
    for(auto y:se){
        ans.push_back(y);
    }
    return ans;
    }
};
10108