Number of Orders in the Backlog LeetCode solution in Python

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# Runtime: 689 ms (Top 70.7%) | Memory: 55.89 MB (Top 32.0%)
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class Solution:
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    def getNumberOfBacklogOrders(self, orders):
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        b, s = [], []
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        heapq.heapify(b)
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        heapq.heapify(s)
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        for p, a, o in orders:
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            if o == 0:
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                heapq.heappush(b, [-p, a])
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            elif o == 1:
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                heapq.heappush(s, [p, a])
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            # Check "good" condition
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            while s and b and s[0][0] <= -b[0][0]:
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                a1, a2 = b[0][1], s[0][1]
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                if a1 > a2:
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                    b[0][1] -= a2
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                    heapq.heappop(s)
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                elif a1 < a2:
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                    s[0][1] -= a1
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                    heapq.heappop(b)
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                else:
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                    heapq.heappop(b)
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                    heapq.heappop(s)
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        count = sum([a for p, a in b]) + sum([a for p, a in s])
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        return count % (10**9 + 7)