<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]:
        
        count = collections.Counter(barcodes)
        heap = [(-val, ind) for ind, val in count.items()]
        heapq.heapify(heap)

        output = []

        while len(heap) > 1:
            first_count, first_ind = heapq.heappop(heap)
            second_count, second_ind = heapq.heappop(heap)
            if output and output[-1] == first_ind:
                output.append(second_ind)
                if second_count + 1 < 0:
                    heapq.heappush(heap, (second_count + 1, second_ind))
                heapq.heappush(heap, (first_count, first_ind))

            else:
                output.append(first_ind)
                if first_count + 1 < 0:
                    heapq.heappush(heap, (first_count + 1, first_ind))
                heapq.heappush(heap, (second_count, second_ind))

        if heap:
            output.append(heap[0][1])
        return output

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