<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]:
        freqs = collections.defaultdict(int)
        for b in barcodes:
            freqs[b] += 1
        pq = [(-freqs[b], b) for b in freqs]
        heapq.heapify(pq)
        res = []
        while len(pq) > 1:
            _, first = heapq.heappop(pq)
            _, second = heapq.heappop(pq)
            res.append(first)
            res.append(second)
            freqs[first] -= 1
            if freqs[first] > 0:
                heapq.heappush(pq, (-freqs[first], first))
            freqs[second] -= 1
            if freqs[second] > 0:
                heapq.heappush(pq, (-freqs[second], second))    
        if pq:
            res.append(pq[0][1])
        return res   
18900