<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>& bar) {
        unordered_map<int, int> freq;
        for (int& n : bar) {
            freq[n]++;
        }

        priority_queue<pair<int, int>> pq;
        for (auto& it : freq) {
            pq.push(make_pair(it.second, it.first));
        }

        vector<int> res;

        int prev = -1;

        while (!pq.empty()) {
            auto it = pq.top();
            pq.pop();

            if (it.second != prev) {
                res.push_back(it.second);
                if (it.first > 1) {
                    pq.push(make_pair(it.first - 1, it.second));
                }
                prev = it.second;
            } else {
                if (!pq.empty()) {
                    auto i = pq.top();
                    pq.pop();
                    pq.push(it);
                    res.push_back(i.second);
                    if (i.first > 1) {
                        pq.push(make_pair(i.first - 1, i.second));
                    }
                    prev = i.second; 
                }
            }
        }

        return res;
    }
};

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