<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 {
private:
void x(vector<int>nums,vector<int>output,int index,vector<vector<int>>&ans){
    if(index>=nums.size())
    {
        auto it = find(ans.begin(), ans.end(), output);
         if(it==ans.end())
          ans.push_back(output);
        return;
    }
    x(nums,output,index+1,ans);
    int e=nums[index];
    output.push_back(e);
    x(nums,output,index+1,ans);

}
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>>ans;
        vector<int>output;
        int index=0;
        sort(nums.begin(),nums.end());
        x(nums,output,index,ans);
        return ans;
    }
};
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