Mine's similar:
steps=:3 :0
bounds=. _1+0,#y
cnt=. 1+ndx=. 0
while. 1=bounds I. nxt=. ndx+off=. ndx{y do.
y=. (1+off) ndx} y
ndx=. nxt
cnt=. 1+cnt
end.
)
But part 2 was only required another 7 characters, so there's that
(though it also took over 70 times as many steps to complete.)
That said, I should perhaps note that my leaderboard score is 0 - I'm
waiting plenty of time before trying these things.
--
Raul
On Tue, Dec 5, 2017 at 5:05 PM, Ric Sherlock <[email protected]> wrote:
> Did anyone come up with a nice Jish solution for this?
>
> Mine (see below) ended up being very scalar.
>
>
>
>
>
>
>
>
>
>
>
>
>
> countJumps=: 3 :0
>
> count=. idx=. 0
>
> offsets=. y
>
> while. (0 <: idx) *. (idx < #offsets) do.
>
> count=. count + 1
>
> idx_new=. idx ([ + {) offsets
>
> offsets=. idx (1 + {)`[`]} offsets
>
> idx=. idx_new
>
> end.
>
> count;idx;offsets
>
> )
>
>
> countJumps 0 3 0 1 _3
>
> 5
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