<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:
    vector<vector<int>> ans;
    void fun(int i , int n, vector<int> &v, vector<int> tmp) {
        if(i>=n) {
            ans.push_back(tmp); return ;
        }
        int j = i+1 ;
        while(j<n && v[j] == v[j-1]) j++;
        fun(j,n,v,tmp);
        for(int k = i ; k<j; k++) {
            tmp.push_back(v[k]);
            fun(j,n,v,tmp);
        }

    }
    vector<vector<int>> subsetsWithDup(vector<int>& v) {
        sort(v.begin(), v.end()); ans.clear();
        int n = v.size();
        vector<int> tmp;
        fun(0,n,v,tmp);
        return ans;
    }
};
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