<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 {
 public:
  vector<int> rearrangeBarcodes(vector<int>& barcodes) {
    vector<int> ans(barcodes.size());
    vector<int> count(10001);
    int i = 0;  // ans' index

    for (const int b : barcodes)
      ++count[b];

    const auto maxIt = ranges::max_element(count);
    const int maxNum = maxIt - count.begin();

    auto fillAns = [&](int num) {
      while (count[num]-- > 0) {
        ans[i] = num;
        i = i + 2 < barcodes.size() ? i + 2 : 1;
      }
    };

    fillAns(maxNum);
    for (int num = 1; num < 10001; ++num)
      fillAns(num);

    return ans;
  }
};
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