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183 comments
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amateurs
def is_even(n: int): if n ==0： return True elif n < 0: return is_even(-n) else: return not is_even(n-1)-
here's a constant time solution:
def is_even(n: int): import math return sum(math.floor(abs(math.cos(math.pi/2 * n/i))) for i in range(1, 2 ** 63)) > 0spoiler
i can't imagine how long it'll take to run, my computer took over 3 minutes to compute one value when the upper bound was replaced with 230. but hey, at least it's O(1).
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Nice, how about bitwise & operator?
// n&1 is 1, then odd, else even return (!(n & 1));
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Don't use recursion. Each call will need to allocate a new stack frame which leads to a slower runtime than an iterative approach in pretty much any language.
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TCO baby
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Since when did Python have tail call optimization?
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183 comments