<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;
        unordered_map<int, int> mp;
        for (int i = 0; i < n; i++) {
            mp[barcodes[i]]++;
        }
        priority_queue<pair<int, int>> pq;
        for (auto& it : mp) {
            pq.push({it.second, it.first});
        }
        while ( !pq.empty() && ans.size() < n) {
            pair<int, int> p = pq.top();
            pq.pop();
            ans.push_back(p.second);
            if (!pq.empty()) {
                pair<int, int> q = pq.top();
                pq.pop();
                ans.push_back(q.second);
                q.first -= 1;
                if (q.first > 0) {
                    pq.push(q);
                }
            }
                 p.first -= 1;
                if (p.first > 0) {
                    pq.push(p);
                }

        }
        return ans;
    }
};
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