<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 solve(vector<int>&nums , int i  ,set<vector<int>>&ans , vector<int>&temp ){
        if( i >= nums.size() ){
            ans.insert(temp) ; 
             return  ; 
        }

        temp.push_back(nums[i]); 
        solve(nums , i+1 , ans  , temp ) ; 
        temp.pop_back() ; 
         solve(nums , i+1 , ans  , temp ) ; 

    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {

        set<vector<int>>temp ; 
        vector<vector<int>>ans  ;
        vector<int>rough ;  
         int i = 0 ; 
  
  sort(nums.begin() , nums.end()) ;
        solve(nums , i , temp  , rough ) ; 

        for(auto it : temp){
             ans.push_back(it) ; 
        }

        return ans  ; 
    }
};
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