<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 getans(vector<int>&nums,int index,  int n, vector<int>temp, vector<vector<int>>&ans){
        if(index==n){
            ans.push_back(temp);
            return;
        }
            //pick 
           temp.push_back(nums[index]);
           getans(nums,index+1, n , temp, ans);
           temp.pop_back();
           
        
   // not pick (agar ham kisi element ko pehle pick nhi kr rhe to aage v nhi krenge , watch learnyard playlist of recursion episode number 10)
        while(index+1<n && nums[index]==nums[index+1])
            index++;
           getans(nums,index+1,n, temp,ans);
           
           
          
           
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>>ans;
        vector<int>temp;
        int n=nums.size();
        int index=0;
        sort(nums.begin(), nums.end());
        getans(nums, index,  n , temp, ans);
        return ans; 
    }
};

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