<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) {
    unordered_map<int, int> mp;
    int mostFreq = 0;
    for (auto &b:barcodes) {
      if (++mp[b] > mp[mostFreq])
        mostFreq = b;
    }
    vector<int> ans(barcodes.size(), 0);
    // cout << barcodes.size() << endl;
    int idx = 0;
    while (mp[mostFreq]) {
      ans[idx] = mostFreq;
      // cout << idx << endl;
      mp[mostFreq]--;
      idx += 2;
      if (idx >= ans.size())
        idx = 1;
    }
    for (auto &m:mp) {
      while (m.second) {
        if (idx >= ans.size())
          idx = 1;
        ans[idx] = m.first;
        m.second--;
        idx += 2;
      }
    }
    return ans;
  }
};
44200