Collision resolution in Java HashMap

Java HashMap uses put method to insert the K/V pair in HashMap. Lets say I have used put method and now HashMap<Integer, Integer> has one entry with key as 10 and value as 17.

If I insert 10,20 in this HashMap it simply replaces the the previous entry with this entry due to collision because of same key 10.

If the key collides HashMap replaces the old K/V pair with the new K/V pair.

So my question is when does the HashMap use Chaining collision resolution technique?

Why it did not form a linkedlist with key as 10 and value as 17,20?

Solution 1:

When you insert the pair (10, 17) and then (10, 20), there is technically no collision involved. You are just replacing the old value with the new value for a given key 10 (since in both cases, 10 is equal to 10 and also the hash code for 10 is always 10).

Collision happens when multiple keys hash to the same bucket. In that case, you need to make sure that you can distinguish between those keys. Chaining collision resolution is one of those techniques which is used for this.

As an example, let's suppose that two strings "abra ka dabra" and "wave my wand" yield hash codes 100 and 200 respectively. Assuming the total array size is 10, both of them end up in the same bucket (100 % 10 and 200 % 10). Chaining ensures that whenever you do map.get( "abra ka dabra" );, you end up with the correct value associated with the key. In the case of hash map in Java, this is done by using the equals method.

Solution 2:

In a HashMap the key is an object, that contains hashCode() and equals(Object) methods.

When you insert a new entry into the Map, it checks whether the hashCode is already known. Then, it will iterate through all objects with this hashcode, and test their equality with .equals(). If an equal object is found, the new value replaces the old one. If not, it will create a new entry in the map.

Usually, talking about maps, you use collision when two objects have the same hashCode but they are different. They are internally stored in a list.

Solution 3:

It could have formed a linked list, indeed. It's just that Map contract requires it to replace the entry:

V put(K key, V value)

Associates the specified value with the specified key in this map (optional operation). If the map previously contained a mapping for the key, the old value is replaced by the specified value. (A map m is said to contain a mapping for a key k if and only if m.containsKey(k) would return true.)

http://docs.oracle.com/javase/6/docs/api/java/util/Map.html

For a map to store lists of values, it'd need to be a Multimap. Here's Google's: http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/Multimap.html

A collection similar to a Map, but which may associate multiple values with a single key. If you call put(K, V) twice, with the same key but different values, the multimap contains mappings from the key to both values.

Edit: Collision resolution

That's a bit different. A collision happens when two different keys happen to have the same hash code, or two keys with different hash codes happen to map into the same bucket in the underlying array.

Consider HashMap's source (bits and pieces removed):

public V put(K key, V value) {
    int hash = hash(key.hashCode());
    int i = indexFor(hash, table.length);
    // i is the index where we want to insert the new element
    addEntry(hash, key, value, i);
    return null;
}

void addEntry(int hash, K key, V value, int bucketIndex) {
    // take the entry that's already in that bucket
    Entry<K,V> e = table[bucketIndex];
    // and create a new one that points to the old one = linked list
    table[bucketIndex] = new Entry<>(hash, key, value, e);
}

For those who are curious how the Entry class in HashMap comes to behave like a list, it turns out that HashMap defines its own static Entry class which implements Map.Entry. You can see for yourself by viewing the source code:

GrepCode for HashMap

Solution 4:

First of all, you have got the concept of hashing a little wrong and it has been rectified by @Sanjay.

And yes, Java indeed implement a collision resolution technique. When two keys get hashed to a same value (as the internal array used is finite in size and at some point the hashcode() method will return same hash value for two different keys) at this time, a linked list is formed at the bucket location where all the informations are entered as an Map.Entry object that contains a key-value pair. Accessing an object via a key will at worst require O(n) if the entry in present in such a lists. Comparison between the key you passed with each key in such list will be done by the equals() method.

Although, from Java 8 , the linked lists are replaced with trees (O(log n))