<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:
    set<vector<int>> ans;
    void backtracking(int cur_idx, vector<int> &nums, vector<int> &vec, vector<bool> &used) {
        if (cur_idx == nums.size()) {
            ans.insert(vec);
            return;
        }

        // if (cur_idx > 0 && nums[cur_idx] == nums[cur_idx - 1] && !used[cur_idx-1]) return;

        backtracking(cur_idx+1, nums, vec, used);
        vec.push_back(nums[cur_idx]);
        used[cur_idx] = true;
        backtracking(cur_idx+1, nums, vec, used);
        vec.pop_back();
        used[cur_idx] = false;
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<int> vec;
        sort(nums.begin(), nums.end());
        vector<bool> used(nums.size(), false);
        backtracking(0, nums, vec, used);
        vector<vector<int>> res(ans.begin(), ans.end());
        return res;
    }
};
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