Minimum Number of Operations to Convert Time LeetCode solution in Python

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class Solution:
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    def convertTime(self, current: str, correct: str) -> int:
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        def get_time(t):
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            hh, mm = t.split(":")
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            return int(hh) * 60 + int(mm)
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        current, correct = get_time(current), get_time(correct)
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        operations = 0
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        diff = correct - current
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        for mins in [60, 15, 5, 1]:
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            quotient, remainder = divmod(diff, mins)
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            operations += quotient
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            diff = remainder
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        return operations