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First solution in Clear category for Striped Words by JnkDog
VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"
PUNCTUATION = ",.!?"
def checkio(text):
text = text.upper()
for c in PUNCTUATION:
text = text.replace(c, " ")
for c in VOWELS:
text = text.replace(c, "v" )
for c in CONSONANTS:
text = text.replace(c, "c" )
words = text.split()
count = 0
for word in words:
if len(word) > 1 and word.isalpha():
if word.find("cc") == -1 and word.find("vv") == -1 :
count += 1
return count
#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"
March 14, 2019
