Question

Here is my issue, lets say I have a std::vector with ints in it.

let’s say it has 50,90,40,90,80,60,80.

I know I need to remove the second, fifth and third elements. I don’t necessarily always know the order of elements to remove, nor how many. The issue is by erasing an element, this changes the index of the other elements. Therefore, how could I erase these and compensate for the index change. (sorting then linearly erasing with an offset is not an option)

Thanks

Answer 1

I am offering several methods:

1. A fast method that does not retain the original order of the elements:

Assign the current last element of the vector to the element to erase, then erase the last element. This will avoid big moves and all indexes except the last will remain constant. If you start erasing from the back, all precomputed indexes will be correct.

void quickDelete( int idx )
{
  vec[idx] = vec.back();
  vec.pop_back();
}

I see this essentially is a hand-coded version of the erase-remove idiom pointed out by Klaim …

2. A slower method that retains the original order of the elements:

Step 1: Mark all vector elements to be deleted, i.e. with a special value. This has O(|indexes to delete|).

Step 2: Erase all marked elements using v.erase( remove (v.begin(), v.end(), special_value), v.end() );. This has O(|vector v|).

The total run time is thus O(|vector v|), assuming the index list is shorter than the vector.

3. Another slower method that retains the original order of the elements:

Use a predicate and remove if as described in https://stackoverflow.com/a/3487742/280314 . To make this efficient and respecting the requirement of
not “sorting then linearly erasing with an offset”, my idea is to implement the predicate using a hash table and adjust the indexes stored in the hash table as the deletion proceeds on returning true, as Klaim suggested.