Why is the worst case complexity of std::unordered_multiset insert is linear? I understand why is that the case for std::unordered_set (you have to check that inserted value is not in the set) but for multiset I don't get it. Am I missing something obvious?
Complexity of std::unordered_multiset insert
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The worst case complexity for
std::unordered_multiset::insert()is linear because:For example, consider the case where
5,13, and13are inserted into anunordered_multisetthat has4buckets, andunordered_multiset::key_eq(5, 13)returnsfalse. In this case,unordered_multiset::hash_function(5)returns different hash codes for both5and13. Despite having different hash codes, these elements may still be inserted into the same bucket. If the hash function for an integer returns the integer itself, and the bucket index is the result of the hash code modulus the number of buckets, then:5is hashed to5, and with4buckets, it is placed in bucket1.13is hashed to13, and with4buckets, it is placed into bucket1as well.While
unordered_set::insert()checks to prevent duplicates during insertion,unordered_multiset::insert()identifies where to insert the element for equivalent-key grouping. In the worst case, the bucket contains[5, 13]when inserting the final13, and upon iterating over all elements, the bucket contains[5, 13, 13]. As iteration over all elements occurs, the complexity is linear insize().It is worth noting that a rehash may occur during
unordered_multiset::insert(), andunordered_multiset::rehash()is specified as having a complexity with an average case linear insize()and the worst case is quadratic. During a rehash, all elements in the original hash table are iterated over and inserted into a new hash table. As iteration has a complexity linear insize(), and as identified above, each insertion has a worse case linear insize(), the resulting worst case isO(size()*size()).