<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:
    void fn(int index, vector<int> current, vector<int>& nums, int n, set<vector<int>>& st) {
        if (index >= n) 
        {
            st.insert(current); 
            return;
        }
        fn(index + 1, current, nums, n, st);
        current.push_back(nums[index]);
        fn(index + 1, current, nums, n, st);
       // current.pop_back();
    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>> xd;
        vector<int> current;
        set<vector<int>> st;

       
        sort(nums.begin(), nums.end());

        
        fn(0, current, nums, nums.size(), st);

        for (const auto& subset : st) {
            xd.push_back(subset);
        }

        return xd;
    }
};
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