<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) {
        map<int,int>m;
        for(auto x:barcodes){
            m[x]++;
        }
        priority_queue<pair<int,int>>pq;
        for(auto x:m){
            pq.push({x.second,x.first});
        }
        vector<int>ans;
        while(pq.size()>1){
            pair<int,int>a,b;
            a=pq.top();pq.pop();
            b=pq.top();pq.pop();
            ans.push_back(a.second);
            ans.push_back(b.second);
            a.first--;
            b.first--;
            if(a.first!=0)pq.push(a);
            if(b.first!=0)pq.push(b);
            // while(b.first--){
            //     ans.push_back(a.second);
            //     ans.push_back(b.second);
            //     a.first--;
            // }
            // if(a.first!=0)pq.push(a);

        }
        if(pq.size()!=0)ans.push_back(pq.top().second);

        return ans;
    }

};
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