Bulb Switcher II LeetCode solution in Python

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# Runtime: 30 ms (Top 97.1%) | Memory: 16.43 MB (Top 14.4%)
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class Solution:
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    def flipLights(self, n: int, m: int) -> int:
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        # Reduce n to at most 3, since any action performed more than 3 times
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        # will result in a pattern that has already been counted
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        n = min(n, 3)
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        if m == 0:
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            return 1
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        elif m == 1:
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            # For m=1, there are only 2 outcomes for n=1, 3 outcomes for n=2, and 4 outcomes for n=3
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            return [2, 3, 4][n - 1]
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        elif m == 2:
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            # For m=2, there are only 2 outcomes for n=1, 4 outcomes for n=2, and 7 outcomes for n=3
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            return [2, 4, 7][n - 1]
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        else:
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            # For m>=3, there are only 2 outcomes for n=1, 4 outcomes for n=2, and 8 outcomes for n=3
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            return [2, 4, 8][n - 1]