It looks like @msp729's translation from 2 years ago fixed this.

This test case will lead to 1 Mln different solutions!!! It will be too long to compare the results.

Because there's no path

There's onle a single point ['B'].length === 1

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')

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

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...

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.

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)

Indeed... Thank you for pointing this out.

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

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

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

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

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

@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).