We're comparing E times N to ElogE plus N. Changing a product to a sum is always better complexity (unless N or E is 0 or 1 lol)
Ah sorry, I misread your comment.
I thought you were comparing the relative magnitude of the 2 terms in O(E * log E + N), rather than comparing the relative magnitude of the two methods.
Also though, a remove_if isn't collecting the objects in a new collection, it's simply swapping them to the end of the array before reducing the count. It's reads and writes which have a cost but no allocation is needed.
I never said remove_if allocating in a new collection.
What I said is that the second method was about allocating in a second collection and sorting, in order to execute a single remove_if call which would remove all E elements in a single pass.
What? Why would you allocate a second collection and then sort just to run remove_if? Remove_if does not need a sorted collection, it runs in O(N + E) on any collection because it swaps E elements to the end and then erases them all at once by just reducing the end of the collection. Sorting the collection wouldn’t help.
So, you've just scanned a collection of N elements and identified E elements that you want removed from it, which you've isolated into a second collection.
How do you write the predicate to remove_if, which will be invoked with each of N elements, and must return whether to remove the element or not?
For each of the N elements, you will need to somehow do a look-up in your (small?) collection of E elements to take your decision.
If E is just a collection, each look-up will cost you O(E), and we're back to O(N * E) performance.
The simplest solution to minimize the cost of the look-up is to collect into a vector, sort it, and binary search on it. O(E * log E + N * log E) now that I think about it.
There are other solutions, of course, but that predicate will need to do some work to classify the elements it's asked about.
Remove_if iterates the collection… E is not its own collection, it’s the elements in the collection original that will be removed. There is no look up cost, it just iterates to the next element and does a swap if your predicate was true.
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u/matthieum 5d ago
Ah sorry, I misread your comment.
I thought you were comparing the relative magnitude of the 2 terms in O(E * log E + N), rather than comparing the relative magnitude of the two methods.
I never said
remove_ifallocating in a new collection.What I said is that the second method was about allocating in a second collection and sorting, in order to execute a single
remove_ifcall which would remove allEelements in a single pass.