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    That's great, many thanks for the reply. I have raised an issue in the test runner GitHub repo.

    Many thanks again, and thanks again for the kata.
    Duncan

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    Hi there,

    I've completed the code so that the tests all pass when attempting the Kata. However, I can't submit my solution because there is a deprecated API being used in the actual test cases.

    Passed: 3 Failed: 0 Errors: 1

    I see:

    STDERR:
    Note: /home/codewarrior/src/main/java/KataTests.java uses or overrides a deprecated API.
    Note: Recompile with -Xlint:deprecation for details.

    As this is stopping this Kata from being completed, would it be possible to fix this up please?

    Many thanks,
    Duncan

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    ==== EDIT =====

    I've completed the kata and my assumptions below about the instructions seemed to be correct :)

    Hi there,

    I was wondering if having completed the Kata you could maybe help me to understand a part of the rules so that I too can finish it. When it mentions "Every sub-array in the array is independent and is only one element for the progress through the array.", that's fine, so if the next element I'm processing is not a number X but rather a subarray, then I will process this subarray with the usual rules.

    It does not however say what to do after processing it - rather, the only instructions given relate to what to do after finding a number X. I'm presuming that you add the processed subarray as a new element in your list and then move onto the next element in the array being processed? But I'm not sure, so any advice welcome thanks!

    Additionally, although it doesn't explicitly mention it in the instructions, I'm presuming that each run results in using the array created in the previous run as its input and it generates a brand new array from that run. And I'm also presuming from seeing examples that when switching directions per run, you also need to switch the direction by which you create a new array i.e. on the second run through, the first element processed (i.e. the right-most element in the original array for this run) will actually be the last element of a new array generated in this run - the walk-through example seems to suggest that rather than explicitly saying it.

    Many thanks for any light you are able to shed on these questions,
    Duncan