Thought I'll add a time complexity analysis since I couldn't find one posted here(or maybe I've missed it):

If we assume T(n) is the time complexity for solving the problem of size n(n * n matrix), we have the following recursion:

T(n) = n * (T(n-1) + n^2)

Basically we have n problems of size n-1 and also n^2 to construct the matrix for each sub problem.

Expanding the above we have:

T(n) = n * T(n-1) + n^3

which gives us

T(n) = n^n

What do you guys think?

Thanks

I did console to track the test #, and also to see your code's output.

You are failing on the fixed test

assert.isTrue(determinant(m7)===88)

const m7 = [
  [2, 4, 5, 3, 1, 2],
  [2, 4, 7, 5, 3, 2],
  [1, 1, 0, 2, 3, 1],
  [1, 3, 9, 0, 3, 2],
  [1, 1, 2, 2, 4, 1],
  [0, 0, 4, 1, 2, 3]
]

where your code returns -88 instead of 88.

I hope that helps!

Ups. I am sorry, but it seems I cannot format it to make it more readable...

Great kata, maybe try to mention the rules of (+ - + -) when having a chain of determinates in the details.

Pretty good Kata! Thanks to author!

Good job, it always feels amazing to finally reach a goal that you set yourself ages ago

I forgot too, but thankfully I realised after reading the formula a couple times

I'm proud of myself for actually solving the problem instead of depending on libraries to do it for me

Yesterday, I mentioned that there was an issue with random tests, and I couldn't express the error accurately at the time. However I am now sure that it is an Overflow Error due to the large matrixes which causes different results for each primitive type. There needs to be a limitation on this!!!!

The issue was related to the returned double variable from .pow() method Thanks for your kindness

I've marked your last comment as spoiler. I think Math.pow uses doubles, and there might be some truncation errors or something.