aha clear solution for Three Words by magda_kudrycka - python coding challenges

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aha solution in Clear category for Three Words by magda_kudrycka

def checkio(words):
    licznik = 0
    for i in words.split():
        if i.isalpha():
            licznik += 1
        else:
            licznik = 0
        if licznik >= 3:
            return True
            
    return False

Oct. 20, 2016

Comments:

maughan.john on Oct. 31, 2016, 10:12 a.m.
Your solution looks very similar to mine. A couple minor variations that are interesting to me are that you didn't use a i.isdigit() test, you just used a straight "else" clause to reset the counter to zero. I actually used an "else" clause on the "for" loop in order to return false, but I see that your solution works as well, as it follows the same logic. If the "for" loop ends without returning True, the next sequential statement is to return False. Interesting! Well done, sir.

magda_kudrycka on Oct. 31, 2016, 7:35 p.m.
the thing is, you don't really need to check if the word contains numbers. The word should be built only with letters, so the wrong input would be not only numbers, but also all of the special chars, like "%" or "*" - literally everything besides letters.

maughan.john on Oct. 31, 2016, 9:24 p.m.
Agreed! I meant to indicate that this was a strength of your code, not a weakness. Your code was more efficient than mine in this regard. Sorry for the misunderstanding!

Strings and Integers

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