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For..else solution in Clear category for Striped Words by a_ntv
VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"
import re
def checkio(text):
ltext = re.split(r'[;,.:?!\s]\s*', text.upper())
qstripped = 0
LETTERS = VOWELS+CONSONANTS
for word in ltext:
lenword = len(word)
if lenword < 2: continue
for i in range(lenword):
if word[i] not in LETTERS: break
if word[i] in VOWELS:
if i+1
April 7, 2020
