Accepted Cheat 1-liner :-D

  • 2

    The judge doesn't check that I'm only using the allowed digits or even digits at all, so I just build a string with the correct length without repeating substrings of length n. Currently it does get accepted (edit: not anymore, see reply below). And hey, if I'm cracking a safe, I might as well cheat.


    def crack_safe(n, k)
      [''].product(*%w(abcdefghij klmnop ABCDEF KLM).map(&:chars)[0, n]).join[0, k**n + n-1]


    def crackSafe(self, n, k):
        chunks = itertools.product(*'abcdefghij klmnop ABCDEF KLM'.split()[:n])
        return ''.join(map(''.join, chunks))[:k**n + n-1]


    Let's say n is 2. Then I concatenate these pairs of characters: "ak", "al", "am", ..., "jn", "jo", "jp". That is, all combinations of a letter from "abcdefghij" with a letter from "klmnop". So my result is "akalam...jnjojp". I just need to cut it off at the right length.

    But how do I know there are no duplicate substrings of length 2? Well, those starting at even indexes are just "ak", "al", etc, all different combinations I used to build the string in the first place. I also can't have a substring at an even index matching a substring at an odd substring, since I use different alphabets ("abcdefghij" and "klmnop") for even and odd indices. Could two substrings at odd indices match? Heh, actually I just noticed that I forgot to think about this. But I think I just get all combinations of a letter from "klmnop" with a letter from "abcdefghij", just not in the normal order.

    When n is 1, 3 or 4, I do the same, except I concatenate singles or triples or quadruples.

    The four alphabet sizes are btw minimized so they're just long enough for all cases.

  • 2

    @StefanPochmann Good observation. I fixed the custom matcher to make this type of solution WA now.

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