Menu
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.
301
HashMap has complexity of O(1) for insertion and lookup.
keySet() method in Java is used to create a set out of the key elements contained in the hash map. It basically returns a set view of the keys or we can create a new set and store the key elements in them. Syntax: hash_map.keySet()
Hashmap put and get operation time complexity is O(1) with assumption that key-value pairs are well distributed across the buckets. It means hashcode implemented is good.
This basically goes for most hash table implementations in most programming languages, as the algorithm itself doesn't really change. If there are no collisions present in the table, you only have to do a single look-up, therefore the running time is O(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!
... iterating through the returned
Setwill take obviouslyO(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!
77
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.
38
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
