<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>> ans;
        vector<int> currentSet;
        
        sort(nums.begin(), nums.end());
        
        function<void(int)> backtrack = [&](int start) {
            ans.push_back(currentSet); 
            
            for (int i = start; i < nums.size(); ++i) {
                if (i > start && nums[i] == nums[i - 1]) {
                    continue;
                }
                
                currentSet.push_back(nums[i]);   
                backtrack(i + 1);                
                currentSet.pop_back();           
            }
        };
        
        backtrack(0); 
        return ans;
    }
};

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