<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 subsetWithDuplicates(vector<int> &nums, int start, vector<int> currSet, vector<vector<int>> &res) {

        int n = nums.size();
        if (start == n) {
            res.push_back(currSet);
            return;
        }

        int i = start+1; int j;
        while (i < n && nums[i] == nums[i-1]) i++;

        j = i;

        subsetWithDuplicates(nums, j, currSet, res);

        for(i = start; i < j; i++) {
            currSet.push_back(nums[i]);
            subsetWithDuplicates(nums, j, currSet, res);
        }

    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {

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

        vector<vector<int>> res;
        vector<int> currSet;

        subsetWithDuplicates(nums, 0, currSet, res);

        return res;
        
    }
};
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