<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) {
        sort(nums.begin(), nums.end());
        dfs(0, vector<int> (), nums);
        return ans;
    }

    void dfs(int i, vector<int> subset, vector<int>& nums){
        if(i == nums.size()){
            ans.push_back(subset);
            return;
        }
            

        subset.push_back(nums[i]);
        dfs(i + 1, subset, nums);

        subset.pop_back();
        int p = i + 1;
        while(p < nums.size() && nums[i] == nums[p])
            ++p;
        dfs(p, subset, nums);
        
        

    }
private:
    vector<vector<int>> ans;
};
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