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First solution in Creative category for Striped Words by vlad.bezden
'''
Our robots are always working to improve their linguistic skills.
For this mission, they research the latin alphabet and its applications.
The alphabet contains both vowel and consonant letters
(yes, we divide the letters).
Vowels -- A E I O U Y
Consonants -- B C D F G H J K L M N P Q R S T V W X Z
You are given a block of text with different words.
These words are separated by white-spaces and punctuation marks.
Numbers are not considered words in this mission
(a mix of letters and digits is not a word either).
You should count the number of words (striped words) where the vowels with
consonants are alternating, that is; words that you count cannot have two
consecutive vowels or consonants.
The words consisting of a single letter are not striped -- do not count those.
Casing is not significant for this mission.
Input: A text as a string (unicode)
Output: A quantity of striped words as an integer.
Precondition:The text contains only ASCII symbols.
0 < len(text) < 105
'''
import re
VOWELS = set('AEIOUY')
CONSONANTS = set('BCDFGHJKLMNPQRSTVWXZ')
def is_striped(word: str) -> bool:
'''Checks if word is striped'''
odd = set(word[::2])
even = set(word[1::2])
return (odd <= VOWELS and even <= CONSONANTS) or \
(odd <= CONSONANTS and even <= VOWELS)
def checkio(text: str) -> int:
return len([word for word in
re.findall(r'[\w]{2,}', text.upper())
if is_striped(word)])
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'
Feb. 2, 2018
