<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> m;
        priority_queue<pair<int,int>> pq;
        vector<int> arr(barcodes.size(),0);
        for(int i=0;i<barcodes.size();i++) m[barcodes[i]]++;
        for(auto x:m) pq.push({x.second,x.first});
        int i=0;
        while(pq.size()>0){
            pair<int, int> p = pq.top();
            pq.pop();
            while (p.first > 0) {
                while (arr[i] != 0)
                    i++;
                arr[i] = p.second;
                p.first--;
                i = i + 2;
                if (i >= arr.size())
                    i = 0;
            }
        }
        return arr;
    }
};
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