I've seen some interesting claims on SO re Java hashmaps and their O(1) lookup time. Can someone explain why this is so? Unless these hashmaps are vastly different from any of the hashing algorithms I was bought up on, there must always exist a dataset that contains collisions.
In which case, the lookup would be O(n) rather than O(1).
Can someone explain whether they are O(1) and, if so, how they achieve this?
A particular feature of a HashMap is that unlike, say, balanced trees, its behavior is probabilistic. In these cases its usually most helpful to talk about complexity in terms of the probability of a worst-case event occurring would be. For a hash map, that of course is the case of a collision with respect to how full the map happens to be. A collision is pretty easy to estimate.
pcollision = n / capacity
So a hash map with even a modest number of elements is pretty likely to experience at least one collision. Big O notation allows us to do something more compelling. Observe that for any arbitrary, fixed constant k.
O(n) = O(k * n)
We can use this feature to improve the performance of the hash map. We could instead think about the probability of at most 2 collisions.
pcollision x 2 = (n / capacity)2
This is much lower. Since the cost of handling one extra collision is irrelevant to Big O performance, we've found a way to improve performance without actually changing the algorithm! We can generalzie this to
pcollision x k = (n / capacity)k
And now we can disregard some arbitrary number of collisions and end up with vanishingly tiny likelihood of more collisions than we are accounting for. You could get the probability to an arbitrarily tiny level by choosing the correct k, all without altering the actual implementation of the algorithm.
We talk about this by saying that the hash-map has O(1) access with high probability
You seem to mix up worst-case behaviour with average-case (expected) runtime. The former is indeed O(n) for hash tables in general (i.e. not using a perfect hashing) but this is rarely relevant in practice.
Any dependable hash table implementation, coupled with a half decent hash, has a retrieval performance of O(1) with a very small factor (2, in fact) in the expected case, within a very narrow margin of variance.
In Java, how HashMap works?
Using hashCode to locate the corresponding bucket [inside buckets container model].
Each bucket is a LinkedList (or a Balanced Red-Black Binary Tree under some conditions starting from Java 8) of items residing in that bucket.
The items are scanned one by one, using equals for comparison.
When adding more items, the HashMap is resized (doubling the size) once a certain load percentage is reached.
So, sometimes it will have to compare against a few items, but generally, it's much closer to O(1) than O(n) / O(log n).
For practical purposes, that's all you should need to know.
Remember that o(1) does not mean that each lookup only examines a single item - it means that the average number of items checked remains constant w.r.t. the number of items in the container. So if it takes on average 4 comparisons to find an item in a container with 100 items, it should also take an average of 4 comparisons to find an item in a container with 10000 items, and for any other number of items (there's always a bit of variance, especially around the points at which the hash table rehashes, and when there's a very small number of items).
So collisions don't prevent the container from having o(1) operations, as long as the average number of keys per bucket remains within a fixed bound.
I know this is an old question, but there's actually a new answer to it.
You're right that a hash map isn't really O(1), strictly speaking, because as the number of elements gets arbitrarily large, eventually you will not be able to search in constant time (and O-notation is defined in terms of numbers that can get arbitrarily large).
But it doesn't follow that the real time complexity is O(n)--because there's no rule that says that the buckets have to be implemented as a linear list.
In fact, Java 8 implements the buckets as TreeMaps once they exceed a threshold, which makes the actual time O(log n).
O(1+n/k) where k is the number of buckets.
If implementation sets k = n/alpha then it is O(1+alpha) = O(1) since alpha is a constant.
If the number of buckets (call it b) is held constant (the usual case), then lookup is actually O(n).
As n gets large, the number of elements in each bucket averages n/b. If collision resolution is done in one of the usual ways (linked list for example), then lookup is O(n/b) = O(n).
The O notation is about what happens when n gets larger and larger. It can be misleading when applied to certain algorithms, and hash tables are a case in point. We choose the number of buckets based on how many elements we're expecting to deal with. When n is about the same size as b, then lookup is roughly constant-time, but we can't call it O(1) because O is defined in terms of a limit as n → ∞.
Elements inside the HashMap are stored as an array of linked list (node), each linked list in the array represents a bucket for unique hash value of one or more keys.
While adding an entry in the HashMap, the hashcode of the key is used to determine the location of the bucket in the array, something like:
location = (arraylength - 1) & keyhashcode
Here the & represents bitwise AND operator.
For example: 100 & "ABC".hashCode() = 64 (location of the bucket for the key "ABC")
During the get operation it uses same way to determine the location of bucket for the key. Under the best case each key has unique hashcode and results in a unique bucket for each key, in this case the get method spends time only to determine the bucket location and retrieving the value which is constant O(1).
Under the worst case, all the keys have same hashcode and stored in same bucket, this results in traversing through the entire list which leads to O(n).
In the case of java 8, the Linked List bucket is replaced with a TreeMap if the size grows to more than 8, this reduces the worst case search efficiency to O(log n).
We've established that the standard description of hash table lookups being O(1) refers to the average-case expected time, not the strict worst-case performance. For a hash table resolving collisions with chaining (like Java's hashmap) this is technically O(1+α) with a good hash function, where α is the table's load factor. Still constant as long as the number of objects you're storing is no more than a constant factor larger than the table size.
It's also been explained that strictly speaking it's possible to construct input that requires O(n) lookups for any deterministic hash function. But it's also interesting to consider the worst-case expected time, which is different than average search time. Using chaining this is O(1 + the length of the longest chain), for example Θ(log n / log log n) when α=1.
If you're interested in theoretical ways to achieve constant time expected worst-case lookups, you can read about dynamic perfect hashing which resolves collisions recursively with another hash table!
It is O(1) only if your hashing function is very good. The Java hash table implementation does not protect against bad hash functions.
Whether you need to grow the table when you add items or not is not relevant to the question because it is about lookup time.
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). If there are collisions present, you have to do more than one look-up, which drives down the performance towards O(n).
It depends on the algorithm you choose to avoid collisions. If your implementation uses separate chaining then the worst case scenario happens where every data element is hashed to the same value (poor choice of the hash function for example). In that case, data lookup is no different from a linear search on a linked list i.e. O(n). However, the probability of that happening is negligible and lookups best and average cases remain constant i.e. O(1).
Only in theoretical case, when hashcodes are always different and bucket for every hash code is also different, the O(1) will exist. Otherwise, it is of constant order i.e. on increment of hashmap, its order of search remains constant.
Academics aside, from a practical perspective, HashMaps should be accepted as having an inconsequential performance impact (unless your profiler tells you otherwise.)
Of course the performance of the hashmap will depend based on the quality of the hashCode() function for the given object. However, if the function is implemented such that the possibility of collisions is very low, it will have a very good performance (this is not strictly O(1) in every possible case but it is in most cases).
For example the default implementation in the Oracle JRE is to use a random number (which is stored in the object instance so that it doesn't change - but it also disables biased locking, but that's an other discussion) so the chance of collisions is very low.
I was going through Rehashing process in hashmap or hashtable
which explains the beautiful concept of
HashMap rehashing.
I have a question about it.
Consider a case where initial capacity of HashMap is 16 and load factor is 0.75. Now, 12 elements are added to the HashMap, but they all end up in the same bucket due to a poor hashCode implementation. The other 15 buckets do not contain any elements. Will the HashMap rehash?
Actually, at least with the current implementation of HashMap, collisions don't affect rehashing. See, for example, lines 661-662 at the end of HashMap.putVal, which simply check
if (++size > threshold)
resize();
where threshold is (capacity * load factor), cached in a field for speed (avoiding a floating-point operation).
HashMap deals with overfull buckets by converting the buckets from linked lists to balanced trees.
Suppose I need to store 1000 objects in Hashset, is it better that I have 1000 buckets containing each object( by generating unique value for hashcode for each object) or have 10 buckets roughly containing 100 objects?
1 advantage of having unique bucket is that I can save execution cycle on calling equals() method?
Why is it important to have set number of buckets and distribute the objects amoung them as evenly as possible?
What should be the ideal object to bucket ratio?
Why is it important to have set number of buckets and distribute the objects amoung them as evenly as possible?
A HashSet should be able to determine membership in O(1) time on average. From the documentation:
This class offers constant time performance for the basic operations (add, remove, contains and size), assuming the hash function disperses the elements properly among the buckets.
The algorithm a Hashset uses to achieve this is to retrieve the hash code for the object and use this to find the correct bucket. Then it iterates over all the items in the bucket until it finds one that is equal. If the number of items in the bucket is greater than O(1) then lookup will take longer than O(1) time.
In the worst case - if all items hash to the same bucket - it will take O(n) time to determine if an object is in the set.
What should be the ideal object to bucket ratio?
There is a space-time tradeoff here. Increasing the number of buckets decreases the chance of collisions. However it also increases memory requirements. The hash set has two parameters initialCapacity and loadFactor that allow you to adjust how many buckets the HashSet should create. The default load factor is 0.75 and this is fine for most purposes, but if you have special requirements you can choose another value.
More information about these parameters can be found in the documentation for HashMap:
This implementation provides constant-time performance for the basic operations (get and put), assuming the hash function disperses the elements properly among the buckets. Iteration over collection views requires time proportional to the "capacity" of the HashMap instance (the number of buckets) plus its size (the number of key-value mappings). Thus, it's very important not to set the initial capacity too high (or the load factor too low) if iteration performance is important.
An instance of HashMap has two parameters that affect its performance: initial capacity and load factor. The capacity is the number of buckets in the hash table, and the initial capacity is simply the capacity at the time the hash table is created. The load factor is a measure of how full the hash table is allowed to get before its capacity is automatically increased. When the number of entries in the hash table exceeds the product of the load factor and the current capacity, the capacity is roughly doubled by calling the rehash method.
As a general rule, the default load factor (.75) offers a good tradeoff between time and space costs. Higher values decrease the space overhead but increase the lookup cost (reflected in most of the operations of the HashMap class, including get and put). The expected number of entries in the map and its load factor should be taken into account when setting its initial capacity, so as to minimize the number of rehash operations. If the initial capacity is greater than the maximum number of entries divided by the load factor, no rehash operations will ever occur.
Roughly one bucket per element is better for the processor, too many buckets is bad for the memory. Java will start with a small amount of buckets and automatically increase the capacity of your HashSet once it starts filling up, so you don't really need to care unless your application has issues performance and you've identified a hashset as the cause.
If you several elements in each bucket, lookups start taking longer. If you have lots of empty buckets, you're using more memory than you need and iterating over the elements takes longer.
This seems like a premature optimization waiting to happen though - the default constructor is fine in most cases.
Object.hashCode()are of type int, you can only have 2^32 different values that's why you create buckets and distribute objects among them.
Edit: If you are using 2^32 buckets to store 2^32 object then defiantly get operations will give you constant complexity but when you are inserting one by one element to store 2^32 objects then rehashing will perform than means if we are using Object[] as buckets then each time it exceeds the length of array it will create new array with greater size and copy elements into this. this process will increase complexity. That's why we make use of equals and hashcode in ratio and that is done by Hashsets itself by providing better hashing algorithm.
Is the retrieval of entries from HashMap really randomized? Or does it depend on buckets in which the entries are hashed and then these buckets are accessed in some predefined order ? or any other kind of order internally happening?
Answer b - it depends on buckets in which the entries are hashed and then these buckets are accessed in some predefined order
What happens when you call the get method on a HashMap for a non-null key:
The hashCode() method for your key is called.
The hash bucket for that hash is obtained by simply and'ing the hash with the number of buckets - 1 which works because the implementation ensures that the number of buckets is always a power of 2. Because the buckets are simply a java array of HashMap.Entry objects the resulting integer can be used as a plain array index. This is where the O(1) complexity comes from.
The entries in the bucket (a linked list) are iterated until one is found where the hash and the key match the one you want. This is where the O(1) complexity starts to deteriorate towards a worst case of O(n).
So you can see that an efficient use of the HashMap will try to minimise the number of entries that end up together in a bucket. All that matters in that respect is the distribution of the bits in your hashcode from bit 0 to bit N-1 where N is the power of 2 used to calculate the number of buckets. All bits in your hashcode above that limit are masked away and only get used again during the iteration of the bucket list.
Here is my situation. I am using two java.util.HashMap to store some frequently used data in a Java web app running on Tomcat. I know the exact number of entries into each Hashmap. The keys will be strings, and ints respectively.
My question is, what is the best way to set the initial capacity and loadfactor?
Should I set the capacity equal to the number of elements it will have and the load capacity to 1.0? I would like the absolute best performance without using too much memory. I am afraid however, that the table would not fill optimally. With a table of the exact size needed, won't there be key collision, causing a (usually short) scan to find the correct element?
Assuming (and this is a stretch) that the hash function is a simple mod 5 of the integer keys, wouldn't that mean that keys 5, 10, 15 would hit the same bucket and then cause a seek to fill the buckets next to them? Would a larger initial capacity increase performance?
Also, if there is a better datastructure than a hashmap for this, I am completely open to that as well.
In the absence of a perfect hashing function for your data, and assuming that this is really not a micro-optimization of something that really doesn't matter, I would try the following:
Assume the default load capacity (.75) used by HashMap is a good value in most situations. That being the case, you can use it, and set the initial capacity of your HashMap based on your own knowledge of how many items it will hold - set it so that initial-capacity x .75 = number of items (round up).
If it were a larger map, in a situation where high-speed lookup was really critical, I would suggest using some sort of trie rather than a hash map. For long strings, in large maps, you can save space, and some time, by using a more string-oriented data structure, such as a trie.
Assuming that your hash function is "good", the best thing to do is to set the initial size to the expected number of elements, assuming that you can get a good estimate cheaply. It is a good idea to do this because when a HashMap resizes it has to recalculate the hash values for every key in the table.
Leave the load factor at 0.75. The value of 0.75 has been chosen empirically as a good compromise between hash lookup performance and space usage for the primary hash array. As you push the load factor up, the average lookup time will increase significantly.
If you want to dig into the mathematics of hash table behaviour: Donald Knuth (1998). The Art of Computer Programming'. 3: Sorting and Searching (2nd ed.). Addison-Wesley. pp. 513–558. ISBN 0-201-89685-0.
I find it best not to fiddle around with default settings unless I really really need to.
Hotspot does a great job of doing optimizations for you.
In any case; I would use a profiler (Say Netbeans Profiler) to measure the problem first.
We routinely store maps with 10000s of elements and if you have a good equals and hashcode implementation (and strings and Integers do!) this will be better than any load changes you may make.
Assuming (and this is a stretch) that the hash function is a simple mod 5 of the integer keys
It's not. From HashMap.java:
static int hash(int h) {
// This function ensures that hashCodes that differ only by
// constant multiples at each bit position have a bounded
// number of collisions (approximately 8 at default load factor).
h ^= (h >>> 20) ^ (h >>> 12);
return h ^ (h >>> 7) ^ (h >>> 4);
}
I'm not even going to pretend I understand that, but it looks like that's designed to handle just that situation.
Note also that the number of buckets is also always a power of 2, no matter what size you ask for.
Entries are allocated to buckets in a random-like way. So even if you as many buckets as entries, some of the buckets will have collisions.
If you have more buckets, you'll have fewer collisions. However, more buckets means spreading out in memory and therefore slower. Generally a load factor in the range 0.7-0.8 is roughly optimal, so it is probably not worth changing.
As ever, it's probably worth profiling before you get hung up on microtuning these things.