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cv(c) or vc(v) solution in Clear category for Striped Words by ZweiStein
VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"
import re
def checkio(text):
regex = (r"\b([" + VOWELS + "]{1,1}[" + CONSONANTS + "]{1,1})+[" + VOWELS + r"]{0,1}\b|"+
r"\b([" + CONSONANTS + "]{1,1}[" + VOWELS + r"]{1,1})+[" + CONSONANTS + r"]{0,1}\b")
test = re.finditer(regex, text.upper())
return len([i for i in test])
#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"
June 20, 2019
