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)
func rearrangeBarcodes(barcodes []int) []int {
	mx := slices.Max(barcodes)
	cnt := make([]int, mx+1)
	for _, x := range barcodes {
		cnt[x]++
	}
	sort.Slice(barcodes, func(i, j int) bool {
		a, b := barcodes[i], barcodes[j]
		if cnt[a] == cnt[b] {
			return a < b
		}
		return cnt[a] > cnt[b]
	})
	n := len(barcodes)
	ans := make([]int, n)
	for k, j := 0, 0; k < 2; k++ {
		for i := k; i < n; i, j = i+2, j+1 {
			ans[i] = barcodes[j]
		}
	}
	return ans
}