<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 FreqMap:
    def __init__(self, num: int, freq: int):
        self.num = num
        self.freq = freq

    def __lt__(self, other):
        return self.freq > other.freq

    def __eq__(self, other):
        return self.freq == other.freq
        
class Solution:
    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
        result = []
        maxheap = []
        lookup = Counter(barcodes)
        for num, freq in lookup.items():
            heapq.heappush(maxheap, FreqMap(num, freq))
        while maxheap:
            first = heapq.heappop(maxheap)
            if len(result) == 0 or first.num != result[-1]:
                result.append(first.num)
                if first.freq - 1 > 0:
                    heapq.heappush(maxheap, FreqMap(first.num, first.freq - 1))
            else:
                second = heapq.heappop(maxheap)
                result.append(second.num)
                if second.freq - 1 > 0:
                    heapq.heappush(maxheap, FreqMap(second.num, second.freq - 1))
                heapq.heappush(maxheap, first)
        return result
18600