<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 backtrack(vector<vector<int>>& output, vector<int>& curr, vector<int>& nums, int index) {
        if(index == nums.size()) { output.push_back(curr); return; }
        
       
        curr.push_back(nums[index]);
        backtrack(output, curr, nums, index+1);
        curr.pop_back();
        
        int i = nums[index];
        index++;
        while(index < nums.size() && nums[index]==i) { index++;}
        backtrack(output, curr, nums, index);
    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> output;
        vector<int> curr;
        backtrack(output, curr, nums, 0);
        return output;
    }
};
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