<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>

set<vector<int>> ans;
void generate(int i,int n, vector<int>& v,vector<int>& x){
    if (i>=n){
        sort(v.begin(),v.end());
        ans.emplace(v);
        return;
    }
    generate(i+1,n,v,x);
    v.push_back(x[i]);
    generate(i+1,n,v,x);
    v.erase(v.begin()+i);
}
class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<int> b;
        ans.clear();
        generate(0,nums.size(),b,nums);
        vector<vector<int>> ans1;
        for (auto it = ans.begin(); it!=ans.end(); ++it)
        {ans1.push_back(*it);}
        return ans1;
    }
};
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