HashMap get/put complexity

Question

We are used to saying that HashMap get/put operations are O(1). However it depends on the hash implementation. The default object hash is actually the internal address in the JVM heap. Are we sure it is good enough to claim that the get/put are O(1) ?

Available memory is another issue. As I understand from the javadocs, the HashMap load factor should be 0.75. What if we do not have enough memory in JVM and the load factor exceeds the limit ?

So, it looks like O(1) is not guaranteed. Does it make sense or am I missing something ?

Answer 1

It depends on many things. It's usually O(1), with a decent hash which itself is constant time... but you could have a hash which takes a long time to compute, and if there are multiple items in the hash map which return the same hash code, get will have to iterate over them calling equals on each of them to find a match.

In the worst case, a HashMap has an O(n) lookup due to walking through all entries in the same hash bucket (e.g. if they all have the same hash code). Fortunately, that worst case scenario doesn't come up very often in real life, in my experience. So no, O(1) certainly isn't guaranteed - but it's usually what you should assume when considering which algorithms and data structures to use.

In JDK 8, HashMap has been tweaked so that if keys can be compared for ordering, then any densely-populated bucket is implemented as a tree, so that even if there are lots of entries with the same hash code, the complexity is O(log n). That can cause issues if you have a key type where equality and ordering are different, of course.

And yes, if you don't have enough memory for the hash map, you'll be in trouble... but that's going to be true whatever data structure you use.

Answer 2

HashMap operation is dependent factor of hashCode implementation. For the ideal scenario lets say the good hash implementation which provide unique hash code for every object (No hash collision) then the best, worst and average case scenario would be O(1). Let's consider a scenario where a bad implementation of hashCode always returns 1 or such hash which has hash collision. In this case the time complexity would be O(n).

Now coming to the second part of the question about memory, then yes memory constraint would be taken care by JVM.

Answer 3

I agree with:

• the general amortised complexity of O(1)
• a bad hashCode() implementation could result to multiple collisions, which means that in the worst case every object goes to the same bucket, thus O(N) if each bucket is backed by a List.
• since Java 8 HashMap dynamically replaces the Nodes (linked list) used in each bucket with TreeNodes (red-black tree when a list gets bigger than 8 elements) resulting to a worst performance of O(logN).

But, this in NOT the full truth if we want to be 100% precise. The implementation of hashCode(), the type of key Object (immutable/cached or being a Collection) might also affect real complexity in strict terms.

Let's assume the following three cases:

1. HashMap<Integer, V>
2. HashMap<String, V>
3. HashMap<List<E>, V>

Do they have the same complexity? Well, the amortised complexity of the 1st one is, as expected, O(1). But, for the rest, we also need to compute hashCode() of the lookup element, which means we might have to traverse arrays and lists in our algorithm.

Lets assume that the size of all of the above arrays/lists is k. Then, HashMap<String, V> and HashMap<List<E>, V> will have O(k) amortised complexity and similarly, O(k + logN) worst case in Java8.

*Note that using a String key is a more complex case, because it is immutable and Java caches the result of hashCode() in a private variable hash, so it's only computed once.

/** Cache the hash code for the string */
    private int hash; // Default to 0

But, the above is also having its own worst case, because Java's String.hashCode() implementation is checking if hash == 0 before computing hashCode. But hey, there are non-empty Strings that output a hashcode of zero, such as "f5a5a608", see here, in which case memoization might not be helpful.

Answer 4

I'm not sure the default hashcode is the address - I read the OpenJDK source for hashcode generation a while ago, and I remember it being something a bit more complicated. Still not something that guarantees a good distribution, perhaps. However, that is to some extent moot, as few classes you'd use as keys in a hashmap use the default hashcode - they supply their own implementations, which ought to be good.

On top of that, what you may not know (again, this is based in reading source - it's not guaranteed) is that HashMap stirs the hash before using it, to mix entropy from throughout the word into the bottom bits, which is where it's needed for all but the hugest hashmaps. That helps deal with hashes that specifically don't do that themselves, although i can't think of any common cases where you'd see that.

Finally, what happens when the table is overloaded is that it degenerates into a set of parallel linked lists - performance becomes O(n). Specifically, the number of links traversed will on average be half the load factor.

Answer 5

It has already been mentioned that hashmaps are O(n/m) in average, if n is the number of items and m is the size. It has also been mentioned that in principle the whole thing could collapse into a singly linked list with O(n) query time. (This all assumes that calculating the hash is constant time).

However what isn't often mentioned is, that with probability at least 1-1/n (so for 1000 items that's a 99.9% chance) the largest bucket won't be filled more than O(logn)! Hence matching the average complexity of binary search trees. (And the constant is good, a tighter bound is (log n)*(m/n) + O(1)).

All that's required for this theoretical bound is that you use a reasonably good hash function (see Wikipedia: Universal Hashing. It can be as simple as a*x>>m). And of course that the person giving you the values to hash doesn't know how you have chosen your random constants.

TL;DR: With Very High Probability the worst case get/put complexity of a hashmap is O(logn).

Answer 6

In practice, it is O(1), but this actually is a terrible and mathematically non-sense simplification. The O() notation says how the algorithm behaves when the size of the problem tends to infinite. Hashmap get/put works like an O(1) algorithm for a limited size. The limit is fairly large from the computer memory and from the addressing point of view, but far from infinity.

When one says that hashmap get/put is O(1) it should really say that the time needed for the get/put is more or less constant and does not depend on the number of elements in the hashmap so far as far the hashmap can be presented on the actual computing system. If the problem goes beyond that size and we need larger hashmaps then, after a while, certainly the number of the bits describing one element will also increase as we run out of the possible describable different elements. For example, if we used a hashmap to store 32bit numbers and later we increase the problem size so that we will have more than 2^32 bit elements in the hashmap, then the individual elements will be described with more than 32bits.

The number of the bits needed to describe the individual elements is log(N), where N is the maximum number of elements, therefore get and put are really O(log N).

If you compare it with a tree set, which is O(log n) then hash set is O(long(max(n)) and we simply feel that this is O(1), because on a certain implementation max(n) is fixed, does not change (the size of the objects we store measured in bits) and the algorithm calculating the hash code is fast.

Finally, if finding an element in any data structure were O(1) we would create information out of thin air. Having a data structure of n element I can select one element in n different way. With that, I can encode log(n) bit information. If I can encode that in zero bit (that is what O(1) means) then I created an infinitely compressing ZIP algorithm.