What is the time complexity of java.util.HashMap class' keySet() method?

Question

I am trying to implement a plane sweep algorithm and for this I need to know the time complexity of java.util.HashMap class' keySet() method. I suspect that it is O(n log n). Am I correct?

Point of clarification: I am talking about the time complexity of the keySet() method; iterating through the returned Set will take obviously O(n) time.

Solution 1

Getting the keyset is O(1) and cheap. This is because HashMap.keyset() returns the actual KeySet object associated with the HashMap.

The returned Set is not a copy of the keys, but a wrapper for the actual HashMap's state. Indeed, if you update the set you can actually change the HashMap's state; e.g. calling clear() on the set will clear the HashMap!

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... iterating through the returned Set will take obviously O(n) time.

Actually that is not always true:

• It is true for a HashMap is created using new HashMap<>(). The worst case is to have all N keys land in the same hash chain. However if the map has grown naturally, there will still be N entries and O(N) slots in the hash array. Thus iterating the entry set will involve O(N) operations.

• It is false if the HashMap is created with new HashMap<>(capacity) and a singularly bad (too large) capacity estimate. Then it will take O(Cap) + O(N) operations to iterate the entry set. If we treat Cap as a variable, that is O(max(Cap, N)), which could be worse than O(N).

There is an escape clause though. Since capacity is an int in the current HashMap API, the upper bound for Cap is 2³¹. So for really large values of Cap and N, the complexity is O(N).

On the other hand, N is limited by the amount of memory available and in practice you need a heap in the order of 2³⁸ bytes (256GBytes) for N to exceed the largest possible Cap value. And for a map that size, you would be better off using a hashtable implementation tuned for huge maps. Or not using an excessively large capacity estimate!

Solution 2

Surely it would be O(1). All that it is doing is returning a wrapper object on the HashMap.

If you are talking about walking over the keyset, then this is O(n), since each next() call is O(1), and this needs to be performed n times.

Solution 3

This should be doable in O(n) time... A hash map is usually implemented as a large bucket array, the bucket's size is (usually) directly proportional to the size of the hash map. In order to retrieve the key set, the bucket must be iterated through, and for each set item, the key must be retrieved (either through an intermediate collection or an iterator with direct access to the bucket)...

**EDIT: As others have pointed out, the actual keyset() method will run in O(1) time, however, iterating over the keyset or transferring it to a dedicated collection will be an O(n) operation. Not quite sure which one you are looking for **

Comments

[M. Justin]

O(n) is not completely accurate, per my answer (the time complexity of iteration is dependent on both the map size as well as its capacity, though the capacity piece shouldn't change the iteration time complexity unless the map is poorly configured for its data).

[M. Justin]

O(n) is not completely accurate, per my answer (the time complexity of iteration is dependent on both the map size as well as its capacity, though the capacity piece shouldn't change the iteration time complexity unless the map is poorly configured for its data).

[Stephen C]

Duplicate answer. Says what I wrote 9 months ago.

[Stephen C]

@M.Justin - And per my answer ... from 9 months earlier.

[Stephen C]

(Note: the 2nd part of this Answer was originally a separate Answer.)

[M. Justin]

@StephenC I'll have take your word for that, as it looks as though you've deleted that answer and merged it into your other one, and I don't have enough reputation to view deleted answers. I must have missed your answer addressing this point.