<p>In a warehouse, there is a row of barcodes, where the <code>i<sup>th</sup></code> barcode is <code>barcodes[i]</code>.</p>

<p>Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> barcodes = [1,1,1,2,2,2]
<strong>Output:</strong> [2,1,2,1,2,1]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> barcodes = [1,1,1,1,2,2,3,3]
<strong>Output:</strong> [1,3,1,3,1,2,1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= barcodes.length &lt;= 10000</code></li>
	<li><code>1 &lt;= barcodes[i] &lt;= 10000</code></li>
</ul>

class Solution:
    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
        from collections import Counter
        import heapq

        # Count frequency of barcodes
        barcode_counts = Counter(barcodes)

        # Create a max heap based on the frequency of barcodes
        max_heap = [(-count, barcode) for barcode, count in barcode_counts.items()]
        heapq.heapify(max_heap)

        previous_barcode = None
        previous_count = 0
        result = []

        while max_heap or previous_count > 0:
            if max_heap:
                current_count, current_barcode = heapq.heappop(max_heap)
                result.append(current_barcode)
                current_count += 1  # Since it's a max heap, we are decrementing the negative count

                # Push the previous barcode back into the heap if it still has remaining counts
                if previous_count < 0:
                    heapq.heappush(max_heap, (previous_count, previous_barcode))

                # Update previous barcode and count
                previous_barcode = current_barcode
                previous_count = current_count
            else:
                # This case handles the edge scenarios where we only have one barcode left but needs to be placed such that no two adjacent are same.
                break

        return result
19000