<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 subset(vector<int>& nums, vector<int>& ref, vector<vector<int>>& res, int ind, bool flag) {
        // Base case: when we've gone through all elements
        if (ind == nums.size()) {
            res.push_back(ref);
            return;
        }

        // Recursive call without including the current element
        subset(nums, ref, res, ind + 1, false);

        // If we're at the start of a new sequence of duplicates or the element is not a duplicate
        if (ind == 0 || nums[ind] != nums[ind - 1] || flag) {
            // Include the current element
            ref.push_back(nums[ind]);
            subset(nums, ref, res, ind + 1, true);
            ref.pop_back(); // Backtrack
        }
    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        // Sort the array to handle duplicates
        sort(nums.begin(), nums.end());
        vector<int> ref;
        vector<vector<int>> res;
        subset(nums, ref, res, 0, true);
        return res;
    }
};

11768