<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:
    /*

    {}

    {1}     {2}     ...   {3}


    */
    vector<vector<int>> subsetsWithDup(vector<int>& nums) 
    {
        std::sort(nums.begin(), nums.end());
        std::vector<std::vector<int> > ret;
        std::vector<int> init_set;
        ret.push_back(init_set);
        ret.reserve((1 << nums.size()));
        dfs(nums, 0, init_set, ret);
        return ret;
    }

    void dfs(const std::vector<int> &nums, int curr_index, std::vector<int> &curr_set, std::vector<std::vector<int> > &ret_set)
    {
        if (curr_index >= nums.size())
            return;

        int index = curr_index;

        // 2   2   2
        while (index < nums.size())
        {
            curr_set.push_back(nums[index]);
            ret_set.push_back(curr_set);
            dfs(nums, index + 1, curr_set, ret_set);
            curr_set.pop_back();
            int next = index + 1;
            while (next < nums.size() && nums[next] == nums[index])
            {
                next++;
            }
            index = next;
        }
    }
};
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