What is the fastest way to see if an array has two common elements?

(^✠Update Below) Insert the array elements to a std::unordered_set and if the insertion fails, it means you have duplicates.

Something like as follows:

#include <iostream>
#include <vector>
#include <unordered_set>

bool has_duplicates(const std::vector<int>& vec)
{
    std::unordered_set<int> set;
    for (int ele : vec)
        if (const auto [iter, inserted] = set.emplace(ele); !inserted)
            return true; // has duplicates!
    return false;
}

int main()
{
    std::vector<int> vec1{ 1, 2, 3 };
    std::cout << std::boolalpha << has_duplicates(vec1) << '\n'; // false

    std::vector<int> vec2{ 12, 3, 2, 3 };
    std::cout << std::boolalpha << has_duplicates(vec2) << '\n'; // true
}

^✠Update: As discussed in the comments, this can or may not be the fastest solution. In OP's case, as explained in Marcus Müller's answer, anO(N·log(N)) method would be better, which we can achieve by having a sorted array check for dupes.

Here is a quick benchmark that I made for the two cases "UnorderedSetInsertion" and the "ArraySort". Following are the result for GCC 10.3, C++20, O3: