<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 rec(int i,vector<int>nums,set<vector<int>>& ans,vector<int>& temp,int n){
        if(i==n){
            // sort(temp.begin(),temp.end());
            ans.insert(temp);
            return;
        }
        temp.push_back(nums[i]);
        rec(i+1,nums,ans,temp,n);
        temp.pop_back();
        rec(i+1,nums,ans,temp,n);
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        int n=nums.size();
        set<vector<int>>ans;
        vector<int>t;
        sort(nums.begin(),nums.end());
        rec(0,nums,ans,t,n);
        // ans.insert({});
        // for(int i=0;i<n;i++){
        // vector<int>a;
        // for(int j=i;j<n;j++){
        //     a.push_back(nums[j]);
        //     ans.insert(a);
        // }
        // }
        vector<vector<int>>x;
        // x.push_back({});
        for(auto i:ans){
            x.push_back(i);
        }
        return x;
    }
};
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