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Simple plan solution in Clear category for Three Words by yuta0613
def checkio(words):
suc_num = 0
for word in words.split():
#word.isalpha()はwordが文字かどうかを判断し、True/Flaseを返す
#数値計算では True = 1, Flase = 0として処理される。
suc_num = (suc_num + 1)*word.isalpha()
if suc_num == 3: return True
else: return False
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio("Hello World hello") == True, "Hello"
assert checkio("He is 123 man") == False, "123 man"
assert checkio("1 2 3 4") == False, "Digits"
assert checkio("bla bla bla bla") == True, "Bla Bla"
assert checkio("Hi") == False, "Hi"
print("Coding complete? Click 'Check' to review your tests and earn cool rewards!")
June 27, 2018
