Hashcode bucket distribution in java

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.