<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> rearrangedCodes(n);
        unordered_map<int, int> numCount;
        for (int i : barcodes) {
            numCount[i]++;
        }

        priority_queue<pair<int, int>> pq;
        for (auto x : numCount) {
            pq.push({x.second, x.first});
        }

        int mostFrequent = pq.top().second;
        int mostFrequentCount = pq.top().first;
        pq.pop();

        int i = 0;
        while (i < n) {
            rearrangedCodes[i] = mostFrequent;
            mostFrequentCount--;
            if (mostFrequentCount == 0) {
                mostFrequent = pq.top().second;
                mostFrequentCount = pq.top().first;
                pq.pop();
            }

            i += 2;
        }

        // if (mostFrequentCount == 0) {
        //     mostFrequent = pq.top().second;
        //     mostFrequentCount = pq.top().first;
        //     pq.pop();
        // }

        i = 1;
        while (i < n) {
            rearrangedCodes[i] = mostFrequent;
            mostFrequentCount--;
            if (mostFrequentCount == 0) {
                mostFrequent = pq.top().second;
                mostFrequentCount = pq.top().first;
                pq.pop();
            }

            i += 2;
        }

        return rearrangedCodes;
    }
};
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