Memoization via a stream isn't good enough to solve this kata. You'll have to study the link provided in the description.

I tried it and it failed in two of the test cases (I think where one array was empty and the other was null)

group() without parameter corresponds to all the characters matching the regex pattern. Group(1) corresponds to the group 1 in the regex pattern.

With the string abc-def and the regex pattern .-(.) group() returns c-d and group(1) returns d

My solution now fails the Woof-shhh-woofwoof! test case as it should.

Random tests can generate woofless subsequences too

What are the odds of this happening? Unfortunately they seem to be too low so I can't consider the issue resolved. My wrong solution still passes 50000 random tests consistently.

That can't be true. I, for example, have this test

assert.strictEqual(woofDecoder("Shhh! Shhh-shhh!"), "nothing to decode!");

To make sure, I added this in my test cases

  it("given \"Woof-shhh-woofwoof!\"", function() {
  assert.strictEqual(woofDecoder("Woof-shhh-woofwoof!"), "b");
  });

Random tests can generate woofless subsequences too

Is it now resolved?

Considering each subsquence one woof (even if it doesn't contain a woof) wouldn't get you through the tests. If I understood you wrong, please clarify

I changed the beginning of the third item in the "things to keep in mind" list to

3. Sequences consist of hyphen-separated subsequences. Each subsequence can contain either zero or one woof (which, in turn, represents zero or one step in the alphabet traversing).

Also, I highlighted the word "letters" in bold while making some minor edits to make the meaning clearer

2. Justin's audio recorder is not a very good one. Sometimes, it inserts some random sounds that should be ignored. With that in mind, a woof is any string of letters that has a "w", an "o", another "o", and an "f" – in that order but not necessarily consecutively and irrespective of the case. If an input string has no woofs, the function should return "nothing to decode!"

You can re-raise (hopefully, providing additional clarifications) in case you deem it insufficient

Do you think it's unclear? How do you think I can improve it?

Each woof subsequence can contain either zero or one woof (which, in turn, represents zero or one step in the alphabet traversing). For example, in a woof sequence "Woof-shhh-woofwoof!" there are three subsequences: "Woof", "shhh", and "woofwoof". The first one contains a woof, the second one doesn't, the third one also has a woof (not two). It means this woof sequence of two woofs in total stands for "b"

also

a woof is any sequence of letters

The expected output of your example string is nothing to decode! since neither of the subsequences (which are: 'w', 'o', 'o', and 'f') contains a woof

Fixed

I think there's a confusion in the thread. The author's solution generates a geometric distribution, which is not wrong per se but generates a distribution that gives short length most of the time and rarely (but can) generate long lengths. See graphs at: https://en.m.wikipedia.org/wiki/Geometric_distribution

The fact that NunoOliveira and Hobovsky are talking about 'the range' is an indication that there's just a misunderstanding.

Hobovsky, your kumite showcases a common mistake when writing tests generating a uniform distribution, but that's not what happening here. Note that the condition here does not depend on the loop counter.

I don't think it's a problem per se that the tests generates a geometric distribution. It's a common mistake to use Math.random() in the loop condition but it's not what happening here.

I'm closing this issue as the author's code is most likely working as intended.

Again, why does that happen under such circumstances?

Because it wasn't fixed. The issue still remains.