<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:
    def findOriginalArray(self, changed: List[int]) -> List[int]:
        counter = collections.Counter(changed)
        res = []
        for k in counter.keys():
            
            if k == 0:
                # handle zero as special case
                if counter[k] % 2 > 0:
                    return []
                res += [0] * (counter[k] // 2)
                
            elif counter[k] > 0:
                x = k
                
                # walk down the chain
                while x % 2 == 0 and x // 2 in counter:
                    x = x // 2
                    
                # walk up and process all numbers within the chain. mark the counts as 0
                while x in counter:
                    if counter[x] > 0:
                        res += [x] * counter[x]
                        if counter[x+x] < counter[x]:
                            return []
                        counter[x+x] -= counter[x]
                        counter[x] = 0
                    x += x
        return res
403