<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 gen(vector<int>& uniqNums,
                         unordered_map<int, int>& freq,
                         vector<int>& curr, int idx,
                         vector<vector<int>>& res) {
        if (idx == uniqNums.size()) {
            sort(res.begin(), res.end());
            res.push_back(curr);
            return;
        }

        int num = uniqNums[idx];
        int count = freq[num];
        gen(uniqNums, freq, curr, idx + 1, res);
        for (int i = 1; i <= count; ++i) {
            curr.push_back(num);
            gen(uniqNums, freq, curr, idx + 1, res);
        }
        for (int i = 0; i < count; ++i) {
            curr.pop_back();
        }
    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        unordered_map<int, int> freq;
        vector<int> uniqNums;
        for (int n : nums) {
            freq[n]++;
        }
        for (auto it : freq) {
            uniqNums.push_back(it.first);
        }

        vector<vector<int>> res;
        vector<int> curr;

        gen(uniqNums, freq, curr, 0, res);

        return res;
    }
};

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