break else break creative solution for Keywords Finder by veky - python coding challenges

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break else break solution in Creative category for Keywords Finder by veky

def checkio(text, words):
    hi = [set(range(i, i + len(word))) for word in words.lower().split()
        for i in range(len(text)) if text.lower()[i:].startswith(word)]
    while True:
        for s, t in __import__("itertools").permutations(hi, 2):
            if s & t: hi.remove(s); hi.remove(t); hi.append(s | t); break
        else: break
    t = list(text)
    for s in sorted(hi, key=min, reverse=True):
        a, b = min(s), max(s) + 1; t[b:b] = ""; t[a:a] = ""
    return "".join(t)

March 12, 2014

Comments:

nickie on March 19, 2014, 8:13 a.m.
<p>The "break else break" part merits <a href="http://docs.python.org/3/reference/compound_stmts.html#the-for-statement">some further study</a> if you don't know that Python has an else clause for loops (as I didn't). Until then, I thought that somebody hacked Veky's account and published this puzzle for him. And then, he hacked mine and published a Veky-style puzzle from there... :-)</p>

veky on March 19, 2014, 2:56 p.m.
<p>That's exactly why I wrote it this way. Guido hates flags ("keeping your flow control in logical variables"), and loop-else is very handy in eliminating them. ;-)</p>

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