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ThreeWords solution in Clear category for Three Words by Adrian_W
def isWord(word):
for i in range(0, len(word)):
if(word[i]>='1' and word[i]<='9'):
return 0
return 1
def substr(word, od, do):
output=''
for i in range(od, do):
output+=word[i]
return output
def checkio(words):
last=0
wordss=0
while words.find(' ')!=-1 :
index=words.find(' ')
print (index, words)
if isWord(substr(words, 0, index))==1:
last=index
wordss+=1
words=substr(words,index+1, len(words))
else:
last=index
wordss=0
words=substr(words,index+1, len(words))
if wordss==3:
return True
if isWord(words)==1:
wordss+=1
if wordss==3:
return True
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"
Oct. 7, 2016
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