<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>

#include <vector>
#include <unordered_map>
#include <queue>

using namespace std;

class Solution {
public:
    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
        priority_queue<pair<int,int>> freq_and_code;     // max_heap
        {
            vector<int> barcode_count (10001);
            for (int& barcode: barcodes) barcode_count[barcode]++;
            for (int i = 0; i < 10001; i++) {
                if (barcode_count[i]) freq_and_code.push({barcode_count[i], i});
            }
        }

        int pos = 0;
        while (!freq_and_code.empty()) {
            pair<int,int>const& p = freq_and_code.top();
            for (int i = 0; i < p.first; i++) {
                barcodes[pos] = p.second;
                pos += 2;
                if (pos >= barcodes.size()) pos = 1;
            }
            freq_and_code.pop();
        }

        return barcodes;
    }
};
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