<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) {
        int n = barcodes.size();
        vector<int> ans(n, 0);
        
        unordered_map<int,int> freq;

        for(int &barcode : barcodes)
            freq[barcode]++;
        
        priority_queue<pair<int,int>> maxHeap;
        for(auto &[key, value] : freq)
            maxHeap.push({value, key});
        
        int index = 0, key = 0, value = 0;
        while(!maxHeap.empty()){
            key = maxHeap.top().second;
            value = maxHeap.top().first;
            maxHeap.pop();
            while(value != 0){
                if(index >= n) index = 1;
                ans[index] = key;
                value--;
                index += 2;
            }
        }
        return ans;
    }
};
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