2224. Minimum Number of Operations to Convert Time
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Difficulty: easy Acceptance: 65% Topics: String, Greedy
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Reference solution (spoiler · python)
# Time: O(1)
# Space: O(1)
# greedy
class Solution(object):
def convertTime(self, current, correct):
"""
:type current: str
:type correct: str
:rtype: int
"""
OPS = (60, 15, 5, 1)
diff = (int(correct[:2])*60+int(correct[3:]))-(int(current[:2])*60+int(current[3:]))
result = 0
for x in OPS:
q, diff = divmod(diff, x)
result += q
return result
Solution from kamyu104/LeetCode-Solutions · MIT
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