<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>res(n);
        unordered_map<int, int>freq;
        
        for(auto bar : barcodes){
            freq[bar] ++;
        }
        
        priority_queue<pair<int, int>>pq;
        for(auto &f : freq){
            pq.push({f.second, f.first});
        }

        vector<int>res(n);
        int idx = 0;

        while(!pq.empty()){
            auto[cnt, bar] = pq.top();
            pq.pop();

            while(cnt > 0 && idx < n){
                res[idx] = bar;
                idx += 2;
                cnt--;
            }

                if(cnt > 0 && idx >= n){
                    idx = 1;
                    while(cnt > 0){
                        res[idx] = bar;
                        idx += 2;
                        cnt--;
                    }
                }
        }
        return res;
    }
};
45900