<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>


#pragma GCC optimize("Ofast","inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native","f16c")
static const auto fast = []() {ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); return 0; } ();
class Solution {
public:
    constexpr vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> subsets = {{}};

        int subsetSize = 0;

        for (int i = 0; i < nums.size(); i++) {
            int startingIndex =
                (i >= 1 && nums[i] == nums[i - 1]) ? subsetSize : 0;
            subsetSize = subsets.size();
            for (int j = startingIndex; j < subsetSize; j++) {
                vector<int> currentSubset = subsets[j];
                currentSubset.push_back(nums[i]);
                subsets.push_back(currentSubset);
            }
        }
        return subsets;
    }
};
8966