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First solution in Clear category for Striped Words by andriutwich
import re
VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"
def checkio(text):
words = [word.strip() for word in re.split(r'\s|,|\.|\?', text)]
def is_valid(word):
if not word.isalpha() or len(word) <= 1:
return False
return (
all((char.upper() in VOWELS) for char in word[::2]) \
and \
all((char.upper() in CONSONANTS) for char in word[1::2]) \
) \
or (\
all((char.upper() in CONSONANTS) for char in word[::2]) \
and \
all((char.upper() in VOWELS) for char in word[1::2]) \
)
ans = sum(is_valid(word) for word in words)
return ans
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio(u"My name is ...") == 3, "All words are striped"
assert checkio(u"Hello world") == 0, "No one"
assert checkio(u"A quantity of striped words.") == 1, "Only of"
assert checkio(u"Dog,cat,mouse,bird.Human.") == 3, "Dog, cat and human"
Jan. 12, 2017
