<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>> ans;
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<int>temp;
        vector<vector<int>>temp1;
        sort(nums.begin(),nums.end());
        help(0,nums,temp);
        return ans;
    }
    void help(int i, vector<int>& nums, vector<int>& temp) {
    // Base case: if we reach the end of the array, add the current subset to the answer
    if (i == nums.size()) {
        ans.push_back(temp);
        return;
    }
    
    // Include the current element in the subset
    temp.push_back(nums[i]);
    help(i + 1, nums, temp);
    
    // Backtrack and remove the current element
    temp.pop_back();
    
    // Skip all duplicates to avoid generating duplicate subsets
    while (i + 1 < nums.size() && nums[i] == nums[i + 1]) {
        i++;
    }
    
    // Explore the subset without including the current element
    help(i + 1, nums, temp);
}

};
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