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    Cannot reproduce, closing ~~

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    thanks, it's fixed now

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    You can make the decision process to be in the regex itself. I'll try to improve error messages so you can get some more info.
    Your solution could benefit from the next regex concepts: groups, groups in replacement, |, &. Let me know any futher problems.

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    Maybe the problem is with the browser..

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    Closing the issue.

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    Unfortunately the Kata and its Test Cases cannot be modified when it is already completed more than 500 times. :(

    Shall I close the issue?

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    Hm I think it's more because there are already have more then 500 users who finished it, so the kata gets locked. Not sure if that should count for translations as well or if the translation also has already reached that threshold...

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    As a long time Java developer, I'll add to Darnor's statment, this is an example of poor design in the Java world.

    Access control and defensive code are taken very seriously in professional environments.

    The mantra is:

    • Never make public what can be package protected.
    • Never make package protected what can be protected.
    • Never make protected what can be private.

    In this case, public access getters methods would be provided, but the members would be private so that they cannot be altered without the 'consent' of the class via a public setter method.

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    This comment is hidden because it contains spoiler information about the solution

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    This comment is hidden because it contains spoiler information about the solution

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    @g964: See the issue above and test number three in Python.

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    Maybe I don't understand very well what you are saying but if a prime isn't a factor of any number of the given array it doesn't show up, otherwise we would have to include all primes.

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    I thank you very much for the feedback. I modified slightly writing:

    [ [p, sum of all ij of I for which p is a prime factor (p positive) of ij] ...]

    Hope it is clearer.

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    Read carefully again the description. The length of the initial array doesn't matter. You have to find first all the prime factors of all the elements of the initial array and then for each prime factors you found to make the sum of the numbers of the initial array that are divised by that factor.