• Custom User Avatar

    This problem still exists, at least for Javascript. Sometimes the random tests catch it and sometimes not.

    Even when my code failed on the following, it still passed the full test suite.

    Test.expect(!oneCharDifference('abcd', 'dcbe'), 'abcd and dcba differ by more than one letter')

  • Custom User Avatar

    Here's the test that I suggest to add (Haskell) which I believe is correct. My solution fails it but still passes current full test suite.

        it "one letter diff, others shuffled" $ do
          aSmallDifference "word" "crow" `shouldBe` False
    
  • Custom User Avatar

    Could you give an example? I guess you don't check letters positions, but you check they are in equal numbers... I can hardly imagine a solution that would check only set of letters and would pass all tests...

  • Custom User Avatar

    Meh. I disagree with many of my Kata rankings. But they are what they are and so I don't lose much sleep over it.

  • Custom User Avatar

    Perhaps we should request to take this kata back to beta for revoting, what do you think? (We've done that before, in rare occasions)

  • Custom User Avatar

    Indeed... Thank you for pointing this out.

  • Custom User Avatar

    My algorithm seems to be able to quickly find several solutions for this... they are just too long to be double-checked manually :-)

  • Default User Avatar

    Long time for this issue to not be fixed. It should not be selectable for Haskell while this is still an issue.

  • Custom User Avatar

    9874 + 1 + 5730 + 980305630 + 563 + 122 + 874963704 + 7 + 9022 + 1 + 9120 + 9819 + 512475704 + 794 + 97920 + 974 + 1 + 270 + 980 + 9120 + 65 + 980 + 2149 + 5730 + 863404 + 2190 + 15903 + 980 + 57349 + 5198034 + 5630400 + 980 + 8633634 + 980 + 2149 + 5300 + 93622 + 903375704 + 980 + 863404 + 65 + 5730 + 980 + 93622 + 30494 + 19 + 980 + 8620 + 65 + 264404 + 79 + 74 + 98030 + 9819 + 480 + 496304 + 36204 + 65 + 20198034 + 15903 + 480 + 419745704 + 803 + 8190 + 655 + 98640 + 50134 + 1 + 91490 + 37404 + 14 + 480 + 80134 + 980 + 20149 + 513 + 86340 + 98640 + 5149 + 863404 + 9819 + 57349 + 8013 + 980 + 93622 + 5200 + 655 + 96 + 980 + 563049 + 980 + 863404 + 9819 + 120394 + 31740 + 980 + 491304 + 65 + 980 + 698034 + 14 + 980 + 93622 + 1441724 + 19 + 980 + 96912 + 48759 + 803 + 90098 + 9013 + 8665 + 655 + 96346 + 14 + 980 + 2149 + 86340 + 56350794 + 794 + 2750 + 980 + 57349 + 5198034 + 8013 + 65 + 980 + 8633634 + 98073 + 50134 + 9819 + 980 + 57304 + 563 + 98073 + 501494 + 133049 + 14 + 980 + 57349 + 5198034 + 30409920 + 980 + 2149 + 65 + 980 + 5730 + 863404 + 980 + 2149 + 93622 + 81314404 + 980 + 563049 + 80139 + 5300 + 19 + 2149 + 65 + 980 + 2149 + 93622 + 122 + 65503 + 98073 + 5730 + 8019 + 96 + 980 + 144749034 + 513 + 655 + 980 + 93622 + 51494 + 794 + 2750 + 4863903 + 14 + 49134 + 3740 + 980 + 863404 + 3049 + 4150 + 15903 + 122 + 48130 + 869 + 5748 + 14 + 98073 + 1557271904 + 917263 + 1 + 36654 + 563 + 98073 + 4150 = 5639304404

  • Custom User Avatar

    I've been away from Codewars from a while, but better late than never -- thanks for your kind words!

  • Default User Avatar

    glad I didn't have to do that because my function is 100% correct :)

  • Custom User Avatar

    @Twilight_Sun, it is actually a valid puzzle:

    There are 10 distinct letters: AEFHILORST, nine distinct leading letters: AFHILORST, and solution without leading zeroes does exist.
    (It's easy to deduce here that E is 0 although no special-case for this deduction is needed to code it).

  • Custom User Avatar

    return nothing

  • Custom User Avatar
  • Custom User Avatar

    ha, have already read that book ;) It's really good and practical although I still have no clue how exactly lens types work together.
    One last thing is to get paid for doing Haskell somehow

  • Loading more items...