<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)
    {
        int const n = barcodes.size(), maxEl = *max_element(barcodes.begin(), barcodes.end());
        vector<int> freq(maxEl + 1);

        for (int num : barcodes) freq[num]++;
        vector<int> ans;     

        priority_queue<pair<int, int>> freqNum; //max
        for (int i = 0; i <= maxEl; i++) if (freq[i]) freqNum.push({freq[i], i});

        while (freqNum.size() > 1)
        {
            pair<int, int> top1 = freqNum.top();
            freqNum.pop();
            pair<int, int> top2 = freqNum.top();
            freqNum.pop();

            top1.first--, top2.first--;
            ans.push_back(top1.second), ans.push_back(top2.second);

            if (top1.first) freqNum.push(top1);
            if (top2.first) freqNum.push(top2);
        }
        if (freqNum.size()) ans.push_back(freqNum.top().second);

        return ans.size() ? ans : barcodes;
    }
};
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