<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:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        sort(nums.begin(),nums.end());
        set<vector<int>> st;
        int n=nums.size();
        int m=pow(2,n);
        vector<vector<int>> res;
        for(int i=0;i<m;i++){
            int j=i;
            vector<int> t;
            int idx=0;
            while(j){
                if((j&1)==1){
                    t.push_back(nums[idx]);
                }
                idx++;
                j>>=1;
            }
            if(st.find(t)==st.end())res.push_back(t);
            st.insert(t);
        }
        return res;
    }
};
9796