<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 {
    // void solve(vector<int> nums, vector<int> output, int index,
    // vector<vector<int> > &ans){
    //     if(index >= nums.size()){
    //         ans.push_back(output);
    //         return;
    //     }

    //     //exclude
    //     solve(nums, output, index+1, ans);

    //     //include
    //     int element = nums[index];
    //     output.push_back(element);
    //     solve(nums, output, index+1, ans);
    // }
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        // vector<vector<int>> ans;
        // vector<int> output;
        // int index = 0;
        // solve(nums, output, index, ans);
        // return ans;
        sort(nums.begin(), nums.end());

        vector<vector<int>> ans;
        int n = nums.size();

        int subset = 1 << n;

        for (int i = 0; i < subset; i++) {
            vector<int> temp;
            for (int j = 0; j < n; j++) {
                if (i & (1 << j)) {
                    temp.push_back(nums[j]);
                }
            }
            ans.push_back(temp);
        }

        map<vector<int>, int> mp;
        for(auto it : ans){
            mp[it]++;
        }

        vector<vector<int>> finalAnswer;
        for(auto it : mp){
            finalAnswer.push_back(it.first);
        }
        return finalAnswer;
    }
};
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