<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> freq;
        for(int x: barcodes){
            freq[x]++;
        }
        if(freq.size()==barcodes.size()){
            return barcodes;
        }
        priority_queue<pair<int,int>> pairs;
        for(pair<int,int> pair:freq){
            pairs.push(make_pair(pair.second,pair.first));
        }
        vector<int> result;
        while(!pairs.empty()){
            auto pair1 = pairs.top();
            pairs.pop();
            result.push_back(pair1.second);
            if(!pairs.empty()){
                auto pair2 = pairs.top();
                pairs.pop();
                result.push_back(pair2.second);
                if(--pair2.first){
                    pairs.push(pair2);
                }
            }
            if(--pair1.first){
                pairs.push(pair1);
            }
        }
        return result;
    }
};
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