<p>An integer array <code>original</code> is transformed into a <strong>doubled</strong> array <code>changed</code> by appending <strong>twice the value</strong> of every element in <code>original</code>, and then randomly <strong>shuffling</strong> the resulting array.</p>

<p>Given an array <code>changed</code>, return <code>original</code><em> if </em><code>changed</code><em> is a <strong>doubled</strong> array. If </em><code>changed</code><em> is not a <strong>doubled</strong> array, return an empty array. The elements in</em> <code>original</code> <em>may be returned in <strong>any</strong> order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> changed = [1,3,4,2,6,8]
<strong>Output:</strong> [1,3,4]
<strong>Explanation:</strong> One possible original array could be [1,3,4]:
- Twice the value of 1 is 1 * 2 = 2.
- Twice the value of 3 is 3 * 2 = 6.
- Twice the value of 4 is 4 * 2 = 8.
Other original arrays could be [4,3,1] or [3,1,4].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> changed = [6,3,0,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> changed is not a doubled array.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> changed = [1]
<strong>Output:</strong> []
<strong>Explanation:</strong> changed is not a doubled array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= changed.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= changed[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findOriginalArray(vector<int>& changed) {        
        int n = changed.size() ; 
        vector<int> ans , empty; 

        if(n % 2 == 1) return empty ; 

        int num[100005] ; 
        memset(num, 0 , sizeof(num)) ;
        for(int x: changed) {
            num[x]++; 
        }

        int len = 0; 

        for(int i = 1; i < 100000; i++) {
            if(num[i]) {
                if(i * 2 > 100000 || len == n/2) {
                    // cout << i << endl;
                    return empty ;
                } 
                int left = n/2 - len;
                if(left < num[i]) {
                    cout << "DM";
                    return empty ;
                }
                len+=num[i] ;
                num[i * 2]-=num[i];
                for(int t = 0 ; t < num[i]; t++) ans.push_back(i); 
            }
        }
        for(int i = 0 ; i < num[0]/2;i++) ans.push_back(0);
        return ans;
    }
};
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