Efficient way to get unique elements from a vector of pairs

Question

I have a vector of pairs of integers that looks somehow like that:

(0, 1)
(1, 9)
(2, 3)
(6, 1)
(4, 0)

I want to extract unique elements from there, so that the result looks as follows:
‍‍0‍, 1, 9, 2, 3, 6, 4 (basically just all numbers without duplicates)

At the moment I'm doing it like that:

std::vector<int> getElements(std::vector<std::pair<int, int>> S) {
    std::vector<int> V;
    for (std::vector<std::pair<int, int>>::iterator i = S.begin(); i != S.end(); i++) {
        if (std::find(V.begin(), V.end(), i->first) == V.end()) {
            V.push_back(i->first);
        }
        if (std::find(V.begin(), V.end(), i->second) == V.end()) {
            V.push_back(i->second);
        }
    }
    return V;
}

Is there any more efficient way to do it?

Answer 1

Your current solution is O(n^2). You can reduce the linear-scan for already seen elements to an amortized O(1) by using std::unordered_set to store the already seen numbers; This will improve your runtime to O(n).

Here is an improved algorithm:

std::vector<int> getElements(std::vector<std::pair<int, int>> S) {
    std::unordered_set<int> ss;
    std::for_each(S.begin(), S.end(), [&ss](const auto& p) {
        ss.insert(p.first);
        ss.insert(p.second);
    });
    return std::vector<int>(ss.begin(), ss.end());
}

See an example Live On Coliru

Answer 2

Is there any more efficient way to do it?

Yes, there is. std::find has O(n) complexity for vector, so repeating it for each element gives you O(n*n) complexity.

A simple alternative is to add every element into std::set. The complexity of building the set is O(n log n).