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I want to use string as a key type in a Dictionary. Like Dictionary<string, string>.

As I understand, if I want to Add a new object, it first calculate a hashcode of the key. So the complexity of Add method boils down to a complexity of String.GetHashCode() method.

What I can't get is how can it be O(1) if to calculate it we still need to iterate through all the characters?

https://learn.microsoft.com/en-us/dotnet/api/system.collections.generic.dictionary-2.add?view=net-9.0

In short, my question is around: does adding a very long string work as quick as adding an empty string, another words does time to insert an element depends on the length of the inserted key (string type)?

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    It's O(1) with respect to the number of elements in dictionary -- it doesn't get more expensive as you add more elements to the dictionary Commented Dec 10, 2024 at 14:36
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    string.GetHashCode is not O(1). Commented Dec 10, 2024 at 14:36
  • Do you mean a Dictionary Lookup cannot be O(1) because that must include a GetHashCode call which is > O(1) ? (trying to understand, what you don't understand, here, sorry) Commented Dec 10, 2024 at 14:37
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    @NikitaAznauryan You can have other types of keys than String in a dictionary. That documentation is for the dictionary only. It does not take into account any cost of calculating the HashCode (which it can not know). Commented Dec 10, 2024 at 14:48
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    While algorithmic complexity is a useful tool, it is useful to keep in mind that it is a simplified model that has some limitations. See the myth of RAM as an example. Commented Dec 10, 2024 at 15:02

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I think you are mixing up the complexities regarding the insertion into the dictionary and the hash calculation of the key of the dictionary.

String.GetHashCode() is O(n) in regards to the length of the string (hash code calculation), but a O(1) step in regards to the overall insertion operation into the dictionary, since you do not have to iterate over the elements of the dictionary.

You wrote "...and then compares it against already existing", there is no need to do that, you would override an existing value anyway. Once the hash is calculated, the key and value can simply be inserted into the dictionary at O(1).

(As pointed out in the comments, one exception is if the capacity of the dictionary increases, then the internal array must be re-allocated and the insertion becomes an O(n) operation.)

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Got it. thank you
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if I want to Add a new object, it first calculate a hashcode of the key and then compares it against already exiting.

No. It does not have to compare with existing keys. If it had to look at keys, insertion could not be O(1), because time would increase as the number of keys increased (not O(N), because it could use more efficient search algorithms, but also not O(1)).

Instead, the hash value also determines the specific storage location within the dictionary, so the dictionary can set the element value directly in the correct location without checking other hashes at all*, and so maintain O(1) insertion. If there is already an existing element stored at this location, depending on your platform/implementation it will either overwrite or throw an exception (C# will throw an ArgumentException)

What I can't get is how can it be O(1) if to calculate it we still need to iterate through all the characters?

Talking about hashcode calculation. From the Dictionary's perspective, this is irrelevant. The dictionary only sees and references the hash. It's not it's job to account for the time to produce the hash, which, depending on the type, may also be O(1) or may be something much worse. Strings do need to check characters, and will be O(n) for hash calculation, but the Dictionary gets to assume the hash.

The Dictionary is only concerned with complexity increase as the number of elements in the dictionary goes up, where an element is a string-as-a-single-unit rather than the individual characters. The complexity of the individual elements is up to their type, and a Dicationary can use anything as a key — not just a string — such that it would be silly to try to talk about Dictionary insertion time if that were part of the conversation, since we'd always have to revert to the worst possible type imaginable.

* Dictionary types are time-efficient, but sometimes space inefficient... but only the for the references rather than the entire objects, and thanks to math not as much as you might expect.

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Just to clarify "If there was an existing value" does not mean HashCode since multiple keys may produce the same HashCode.
@Magnus Trying to add some clarity there.

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