<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>& nums) {
        if(nums.size() == 1 || nums.size() == 0)
            return nums;
        unordered_map<int,int> freq;
        for(auto& it:nums)
            freq[it]++;
        priority_queue<pair<int,int>> q;
        for(auto& it:freq)
            q.push(make_pair(it.second,it.first));
        nums.clear();
        while(q.top().first > 0)
        {
            pair<int,int> p = q.top();
            q.pop();
            nums.push_back(p.second);
            p.first--;
            pair<int,int> p2 = q.top();
            q.pop();
            if(p2.first > 0){
                nums.push_back(p2.second);
                p2.first--;
            }
            q.push(p);
            q.push(p2);
        }
        return nums;
    }
};
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