This kata is a duplicate of https://www.codewars.com/kata/60df63c6ce1e7b0023d4af5c

No, it is that the reference solution is consistently wrong when s == l[0] and len(l) > 1. For example:

l = [1, 2], s = 1

1 should equal 0

The kata has performance requirements, so it should specify input range.

Good question. I'll look into it.

I'm seeing a bug in the random tests, here's an example:

With these inputs: l = [3, -1, 1, -8, -6, 10, 3, 0, -7, 0], and s = 3.

I find these four solutions all summing to 3:

1. sum(l[0:1]) => sum([3]) => 3
2. sum(l[0:3]) => sum([3, -1, 1]) => 3
3. sum(l[6:7]) => sum([3]) => 3
4. sum(l[6:8]) => sum([3, 0]) => 3

The error message says I should be finding 3 not 4 solutions, are solutions that end in zero excluded from the count?

updated as above.

I went with your original premise and updated the description with a hint: somestring.count('')

So in the test cases, there is now:

test.assert_equals(rolling_sum([1,2,3],0),4)

I think I see what you are saying. If we say that sub-arrays must have length > 0 then

rs([],0)

would be 0 instead of 1, which is mathematically incorrect

I don't like the idea of removing s = 0, though - that seems even more "special", requiring a special line of code (in solutions and tests) rather than a change in algorithm. I might have to agree with your original assessment to use the empty sets like the builtin does.

I would argue that logically, we should count at most one way to make an empty set.

A sequence of length n includes a contiguous subsequence of length m occurs n - m + 1 times, why should 0 be a exception?

For example, sum(xs[3:3]) doens't make sense for an array of length 3. By this logic, all of these return 0:

Maybe 3:3 isn't the best notation for this, but there's an empty subsequence after the last item. It can be 3:, 4:4 or whatever, but it's there. A built-in for example:

>>> 'aaa'.count('')
4

the length of the sub-array must be > 0

If zero sums would still be valid, the it's an artificial special case, which isn't good.

specifying that using an empty set counts as only 1 solution, even if it can be derived multiple ways

But it's artificial too, so some solutions that naturally count all empty subsequences would have to handle this case separately, which isn't good.

There's an option of removing s = 0 from the input domain.

I would argue that logically, we should count at most one way to make an empty set.

For example, sum(xs[3:3]) doens't make sense for an array of length 3. By this logic, all of these return 0:

sum(xs[4:4])
sum(xs[4:9999999])
sum(xs[-1:-1])

so there would be an infinite number of ways to sum to 0.

Would you consider this issue better resolved by stating in the problem that the length of the sub-array must be > 0 or by specifying that using an empty set counts as only 1 solution, even if it can be derived multiple ways?

I think it should be the latter because it makes sense that

rolling_sum([],0)

should equal 1, not 0.

It's not specified that the subsequences must be non-empty, so there are 4 ways for xs = [1, 2, 3]: sum(xs[0:0]), sum(xs[1:1]), sum(xs[2:2]), sum(xs[3:3]).

Added 10 smaller random tests

It seems like rolling_sum([1,2,3],0) should return 0 since there are no ways to make a sum of 0 with the integers given. You feel like that is not the proper solution?

The reference solutions fails at s = 0:
test.assert_equals(rolling_sum([1,2,3],0),4) -> 0 should equal 4

Missing edge case tests with s = 0.