<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:
    unordered_map<int, int> freq;

    struct comp {
        unordered_map<int, int>& freq_ref;
        comp(unordered_map<int, int>& freq) : freq_ref(freq) {}
        bool operator()(int x, int y) { return freq_ref[x] < freq_ref[y]; }
    };

    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
        if (barcodes.empty())
            return {};
        for (auto x : barcodes)
            freq[x]++;
        priority_queue<int, vector<int>, comp> pq((comp(freq)));
        for (auto x : freq)
            pq.push(x.first);
        vector<int> res;
        while (!pq.empty()) {
            	if (res.size() >= 1 && res.back() == pq.top()) {
	    int y = pq.top(); pq.pop();
	    if (pq.empty()) return res;
	    int x = pq.top(); pq.pop();
	    freq[x]--;
	    if (freq[x] > 0) pq.push(x);
	    res.push_back(x);
	    pq.push(y);
	} else {
	    int x = pq.top(); pq.pop();
	    freq[x]--;
	    if (freq[x] > 0) pq.push(x);
	    res.push_back(x);
	}

        }
        return res;
    }
};
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