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regex power solution in Speedy category for Striped Words by hwmrocker
VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"
import re
striped = re.compile(r'\b(?:['+CONSONANTS+r']['+VOWELS+r'])+(?:['+CONSONANTS+r'])?\b|\b(?:['+VOWELS+r']['+CONSONANTS+r'])+(?:['+VOWELS+r'])?\b')
def checkio(text):
return len(striped.findall(text.upper()))
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio("My name is ...") == 3, "All words are striped"
assert checkio("Hello world") == 0, "No one"
assert checkio("A quantity of striped words.") == 1, "Only of"
assert checkio("Dog,cat,mouse,bird.Human.") == 3, "Dog, cat and human"
July 24, 2014
