<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 {
private:
    vector<vector<int>> v;
    set<vector<int>> s;
public:
    //  app - 1
    // vector<vector<int>> subsetsWithDup(vector<int>& nums) {
    //     vector<int> t;
    //     sort(nums.begin(), nums.end());
    //     sb(0, nums, t);
    //     for(vector<int> vec:s) {
    //         v.push_back(vec);
    //     }
    //     return v;
    // }

    // void sb(int ind, vector<int>& nums, vector<int>& t) {
    //     if(ind == nums.size()) {
    //         s.insert(t);
    //         return;
    //     }
    //     t.push_back(nums[ind]);
    //     sb(ind+1, nums, t);
    //     t.pop_back();
    //     sb(ind+1, nums, t);
    // }

    //  app - 2
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<int> t;
        sort(nums.begin(), nums.end());
        sb(0, nums, t);
        return v;
    }

    void sb(int ind, vector<int> nums, vector<int> t) {
        v.push_back(t);
        for(int i=ind; i<nums.size(); i++) {
            if(i>ind && nums[i]==nums[i-1])
                continue;
            t.push_back(nums[i]);
            sb(i+1, nums, t);
            t.pop_back();
        }
    }
    
};
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