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First solution in Speedy category for Three Words by likewind
def checkio(words):
words = words.split()
if len(words) < 3:
return False
for i in range(len(words)-2):
if words[i].isalpha() and words[i+1].isalpha() and words[i+2].isalpha():
return True
return False
April 16, 2018