On Thursday, 21 December 2017 at 00:52:29 UTC, Nicholas Wilson
wrote:
On Thursday, 21 December 2017 at 00:23:08 UTC, Steven
Schveighoffer wrote:
I'm assuming here indices is sorted? Because it appears you
expect that in your code. However, I'm going to assume it
isn't sorted at first.
auto sortedIdxs = indices.assumeSorted; // also could be =
It's not going to be as good as hand-written code, complexity
wise, but it's definitely shorter to write :)
-Steve
If indices is sorted with no duplicates and random access then
you can do it in linear time.
Ah yes, I guess sorted and unique as well would be the expected
input. But nice to see the handling of non-sorted indices.
I tried to search for an assumeUnique function as well (the
assumeSorted one was nice to see) but it's not what I thought
it'd be - seems to be more of an assumeUnaliased. And I guess
there's no assumeUniqueElements.
And the filter approach is nice! :) (just need to account for the
last ii++ (when filter comes back in I think one case would be ii
== indices.length and you'd get a range error)
Thanks for the feedback!
