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Comment by phamilton | Hacker News Reader

`..into a new array` is precisely what `reject` does. We're talking about `reject!` which modifies the existing array.

Isn't the simple solution for doing it in place "swap reject element position with current end of array, decrement size of array?" What am i missing?

Wouldn't that change the order of elements?

The only guarantee i saw was that the output array did not contain the rejected elements. (and looking at the reference docs, that's all it says)

It appears order is guaranteed to be maintained but it's not specified :)

No. [1,2,3].reject! { |x| x == 1 } would give you [3,2], rearranging the order of elements. It would work for sets.

maybe there's a finalization step that's necessary to free up memory? This seems like the most straightforward solution though

xg15's comment still makes sense if you replace "copy all ..." with "slide the remaining elements down by the number of removed elements" to match the mutative behavior.

However, a bigger problem is concurrent modification: What happens if your reject! predicate has a side effect which modifies the array? Ideally you'd receive a "ConcurrentModificationException" or the like.

That would still require O(n) space (to store the list of elements to keep or remove) so why not simply use the regular `reject` ?

No, before sliding elements, you set a local variable "dirty" to true. You iterate over all elements and keep the ones that are not rejected by moving them at the end of the current max index. When you code finishes normally, dirty is set to false.

You have an "ensure" block (finally) protecting the code which handles the cases where dirty is true (moving all the remaining elements after the current max). Then you truncate the array.

Because you're saving the O(N) time copy...

iirc, Ruby's array doesn't expose the distinction between length and capacity, but internally it may decide to amortize the cost of such copies. For example, if reject only removes a small % of elements, it may choose not to shrink capacity.

You're not saving the O(N) time copy, just replacing it with an O(N) slide.

(That's assuming you copy at all, instead of simply updating the reference.)

I meant O(N) space, as you said "That would still require O(n) space" - but a slide does not require additional space.

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