14
Last seen 7 years ago
Member for 8 years, 7 months, 11 days
Difficulty Normal
Great to know that set.intersection() can accept any iterable. Thank you!
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Nice solution. But you might make it a couple lines shorter:
put safe_pawn.append(pawn) directly under if gaurdy + gaurdx in pawns.
So you don't need the variable safe anymore.
Also, keeping the safe pawns in a list seems to incur redundant memory usage. Keep a int variable count and += 1 when safe
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Stolen from:
http://www.checkio.org/mission/three-words/publications/LLluma/python-3/first/
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similar but without islice:
def checkio(words):
return any(all(map(unicode.isalpha, trigram)) for trigram in zip(*(words.split()[i:] for i in range(3))))
http://www.checkio.org/mission/three-words/publications/hezhe88/python-27/second/
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If anyone is also wondering how filter function eliminates zeros:
https://docs.python.org/2/library/functions.html#filter
"[item for item in iterable **if item**] if function is None."
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This might be an O(nlogn) solution. Can you do it in O(n)?
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First-therods95
Each time text.count(x) is called, it requires O(n) time. So the total running time will be O(n^2).
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