Efficiently get Vec> from Ref<'a, BTreeSet>

Question

I have a Ref<'a, BTreeSet<T>> and I would like to get the references of its content as Vec<Ref<'a, T>>.
One method to do this is:

fn get_refs<'a, T: Ord>(btree: Ref<'a, BTreeSet<T>>) -> Vec<Ref<'a, T>> {
    let mut result = Vec::new();
    for e in btree.iter() {
        result.push(Ref::map(Ref::clone(&btree), |t| t.get(&e).unwrap()))
    }
    result
}

A running example

Now let n be the size of btree.
Because getting a value from a binary tree takes O(log(n)) and because iterating through a binary tree takes O(n) this method has time complexity O(n log(n)).

Even though doing this in with &'a BTreeSet<T> and &'a T instead of Ref<'a, BTreeSet<T>> and Ref<'a, T> takes O(n) (Because we only need to collect the references iterated in an array). An example of the method using plain references is the following.

fn get_refs<'a, T: Ord>(btree: &'a BTreeSet<T>) -> Vec<&'a T> {
    btree.iter().collect()
}

A running example

My question is:
Given Ref<'a, BTreeSet<T>> is there a way to get Vec<Ref<'a, T>> in O(n) time complexity?

If you care that much about complexity it might be appropriate to have both data structures from the beginning and use the set to get items and to use the vector to iterate them. When you add an item, you do so for both structures, so adding is still amortized O(log N) — Alexey S. Larionov, Nov 20 '20 at 14:10

Aplet123 · Accepted Answer · 2020-11-20T14:39:55.753

The easiest way is to take a reference to a Ref instead of a Ref:

fn get_refs<'a, T>(btree: &'a Ref<BTreeSet<T>>) -> Vec<&'a T> {
    btree.iter().collect()
}

Playground link

This way, the Ref can live longer than the life of your function, which means you can return references that borrow the Ref without dealing with borrowing a temporary.

Note that this means you can just work with &'a BTreeSet<T> instead and not even have to work with Refs, so your original code with references works.

Efficiently get Vec> from Ref<'a, BTreeSet>

1 Answers1