<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>

from collections import Counter

class Solution:
    def findOriginalArray(self, changed: List[int]) -> List[int]:
        # If the length of changed is odd, it cannot be a doubled array
        if len(changed) % 2 != 0:
            return []
        
        # Sort the array
        changed.sort()
        
        # Initialize the counter for the elements
        count = Counter(changed)
        
        original = []
        
        # Iterate through the sorted array
        for num in changed:
            # If the current number is still in the counter
            if count[num] > 0:
                # Check if its double is in the counter
                if count[2 * num] > 0:
                    # Add the original number to the result
                    original.append(num)
                    # Reduce the frequency of both the original number and its double
                    count[num] -= 1
                    count[2 * num] -= 1
                else:
                    # If the double is not in the counter, return empty array
                    return []
        
        # Return the original array
        return original

# Example usage:
solution = Solution()
changed = [1, 3, 4, 2, 6, 8]
print(solution.findOriginalArray(changed))  # Output: [1, 3, 4]
1042