How expensive is comparing two unordered sets for equality?

Question

Given two std::sets, one can simply iterate through both sets simultaneously and compare the elements, resulting in linear complexity. This doesn't work for std::unordered_sets, because the elements may be stored in any order. So how expensive is a == b for std::unordered_set?

Answer 1

Complexity of operator== and operator!=:

Linear complexity in the average case. N² in the worst case, where N is the size of the container.

More details in the standard §23.2.5, point 11:

For unordered_set and unordered_map, the complexity of operator== (i.e., the number of calls to the == operator of the value_type, to the predicate returned by key_equal(), and to the hasher returned by hash_function()) is proportional to N in the average case and to N² in the worst case, where N is a.size().