<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]:
        # int array orig: doubled every element, appended, then permuted

        # suppose we sorted the original array, values
        #  1 ...... M for max M
        # then after doubling we'd expect to see 2*M

        # so for every x==2*y we expect to see a y
        # y < 2*y for y > 0 so if we see a remaining odd
        # not paired with an even then it's no good
        # or if we see an even with no corresponding half

        counts = Counter(changed)

        if counts.get(0, 0) & 1:
            # for every original zero we expect a double
            return []

        uniques = list(counts.keys())
        uniques.sort(reverse=True)
        if uniques[-1] == 0: uniques.pop()

        ans = []
        for u in uniques:
            f = counts[u]
            if u & 1:
                if f: return [] # have odd y, 2*y doesn't exist
            else:
                half = u >> 1
                if counts.get(half, 0) >= f:
                    ans.extend(half for _ in range(f))
                    counts[half] -= f
                else:
                    return []

        # FIXED: handle zero specially
        ans.extend(0 for _ in range(counts.get(0, 0)//2))

        return ans
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