One reason would be because std::set guarantees O(log n) complexity for each operation on the worst case. std::unordered_set has average complexity O(1) and O(s

One reason would be because std::set guarantees O(log n) complexity for each operation on the worst case. std::unordered_set has average complexity O(1) and O(size) for the worst case (it being a hash table) which can be unexpected to debug in those rare cases.