Distant Barcodes
<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>
Greedy; Array; Hash Table; Counting; Sorting; Heap (Priority Queue)
class Solution:
    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
        cnt = Counter(barcodes)
        barcodes.sort(key=lambda x: (-cnt[x], x))
        n = len(barcodes)
        ans = [0] * len(barcodes)
        ans[::2] = barcodes[: (n + 1) // 2]
        ans[1::2] = barcodes[(n + 1) // 2 :]
        return ans
