<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:
    using ip=pair<int,int>;
    vector<int> rearrangeBarcodes(vector<int>& bc) {
        priority_queue<ip> pq;
        sort(bc.begin(),bc.end());
        int i=0,n=bc.size();
        while(i<n){
            int cnt = 1;
            while(i+1<n && bc[i]==bc[i+1]){
                cnt++;
                i++;
            }
            pq.push({cnt,bc[i]});
            i++;
        }
        vector<int> ans;
        while(!pq.empty()){
            ip f = pq.top();
            pq.pop();
            ans.push_back(f.second);
            if(pq.size()){
                ip s = pq.top();
                pq.pop();
                ans.push_back(s.second);
                if(s.first-1)
                    pq.push({s.first-1,s.second});

            }
            if(f.first-1)
                pq.push({f.first-1,f.second});
            
        }
        return ans;
    }
};
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