<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> freq;
        for (auto& code : barcodes) {
            freq[code]++; 
        }   
        priority_queue<pair<int,int>> pq;
        for (auto& [code, count] : freq) {
            pq.push({count, code});
        }
        vector<int> ans;
        while (pq.size() > 1) {
            auto a = pq.top(); pq.pop();
            auto b = pq.top(); pq.pop();
            ans.push_back(a.second);
            ans.push_back(b.second);
            a.first--;
            b.first--;
            if (a.first > 0) pq.push(a); 
            if (b.first > 0) pq.push(b); 
        }
        if (!pq.empty()) ans.push_back(pq.top().second);
        return ans;
    }
};

auto boom = [](){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 0;
}();
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