<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) {
        int n = nums.size();
        vector<vector<int>> ans;
        vector<int> set;
        sort(nums.begin(), nums.end());
        
        function<void(int)> subset = [&](int idx) {
            if (idx >= n) {
                ans.push_back(set);
                return;
            }
            
            set.push_back(nums[idx]);
            subset(idx + 1);
            set.pop_back();
            
            while (idx + 1 < n && nums[idx] == nums[idx + 1]) {
                idx++;
            }
            subset(idx + 1);
        };
        
        subset(0);
        return ans;
    }
};
12183