<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>mp;
        for(auto & barcode: barcodes){
            mp[barcode]++;
        }
        priority_queue<pair<int,int>>pq;

        for(auto &m:mp){
           pq.push({m.second,m.first});
        }
        int n=barcodes.size();
        vector<int>ans(n);
        int index1=0;

        while(!pq.empty()){
            auto curr=pq.top();
            int num=curr.second;
            int cnt=curr.first;
            pq.pop();

            while(cnt--){
                if(index1>=n)index1=1;
                ans[index1]=num;
                index1+=2;
            }
        }

        return ans;

    }
};
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