<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>& b, int pos = 0) {
        
        unordered_map<int, int> m;
        set<pair<int, int>> s;
        for (auto n : b)
            ++m[n];
        
        for (auto it = begin(m); it != end(m); ++it) 
            s.insert({ it->second, it->first });
        
        // rbegin returns a reverse iterator pointing to the last element in the vector
        // for this case, it is the largest one
        for (auto it = s.rbegin(); it != s.rend(); ++it) {
            for (auto cnt = 0; cnt < it->first; ++cnt, pos += 2) {
                if (pos >= b.size()) 
                    pos = 1;
                b[pos] = it->second;
            }
        }
        return b;
    }
};




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