<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) {
        // has to sort by frequency and put the highest freq in even positions
        int n = barcodes.size();
        unordered_map<int, int> mq; // val, freq
        for (int i = 0; i < n; i++) {
            mq[barcodes[i]]++;
        }
        
        std::sort(barcodes.begin(), barcodes.end(), [&](int a, int b) {
            return mq[a] > mq[b]  || (mq[a] == mq[b] && a < b);});
        
        vector<int> result(n, 0);
        int ct = 0;
        for (int i = 0; i < n; i+=2) result[i] = barcodes[ct++];
        for (int i = 1; i < n; i+=2) result[i] = barcodes[ct++];
        return result;
    }
};
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