Distant Barcodes LeetCode solution in Python

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# Runtime: 1287 ms (Top 5.56%) | Memory: 16.2 MB (Top 67.41%)
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import heapq
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class Solution:
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    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
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        barcodes_counter = Counter(barcodes)
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        if len(barcodes_counter) == len(barcodes):
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            return barcodes
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        barcodes_heapq = [(-c, b) for b, c in barcodes_counter.items()]
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        heapq.heapify(barcodes_heapq)
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        idx, prev_count, prev_barcode = 0, 0, 0
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        while barcodes_heapq:
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            (curr_count, curr_barcode) = heapq.heappop(barcodes_heapq)
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            barcodes[idx] = curr_barcode
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            idx += 1
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            curr_count += 1
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            if prev_count:
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                heapq.heappush(barcodes_heapq, (prev_count, prev_barcode))
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            prev_count, prev_barcode = curr_count, curr_barcode
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        return barcodes