<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:
    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
        n = len(barcodes)

        if n <= 2:
            return barcodes

        d = {}
        for a in barcodes:
            if a in d:
                d[a] += 1
            else:
                d[a] = 1

        arr = [[-v,k] for k,v in d.items()]

        l = len(arr)

        heapq.heapify(arr)

        ans = []

        item = heapq.heappop(arr)

        while arr:
            ans.append(item[1])

            item[0] += 1

            if item[0] < 0:
                h = heapq.heappop(arr)
                heapq.heappush(arr,item)
                item = h
            else:
                item = heapq.heappop(arr)
        
        ans.append(item[1])

        return ans
19100