<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]:
        """
        priority queue
        """
        count = [(-freq, key) for key, freq in collections.Counter(barcodes).items()]
        pq = []
        for freqKey in count:
            heappush(pq, freqKey)
        result = []
        while pq:
            freq1, key1 = heappop(pq)
            if pq:
                freq2, key2 = heappop(pq)
            else:
                result.append(key1)
                break
            if result and result[-1]==key1:
                result.append(key2)
                result.append(key1)
            else:
                result.append(key1)
                result.append(key2)
            if freq1+1!=0:
                heappush(pq, (freq1+1, key1))
            if freq2+1!=0:
                heappush(pq, (freq2+1, key2))
        return result
# TC: O(N*log(N))
# SC: O(N)
18800