<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) {
       vector<vector<int>> res;
       unordered_map<string,int> hmap;
       sort(nums.begin(), nums.end());
        int n = nums.size();
        for (int b = 0; b < (1 << n); b++) {
            vector<int> sol;
            for (int i = 0; i < n; i++) {
                if (b & (1 << i))
                    sol.push_back(nums[i]);
            }
            string s = convertString(sol);
            if(hmap.find(s) == hmap.end()){
                hmap[s] = 1;
                res.push_back(sol);
            } 
        }
        return res;
    }

    string convertString(vector<int> sol){
        string s;
        for(auto x : sol)
        s += to_string(x);

        return s;
    }
};
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