CODEWITHSUNDEEP
data-structureshashmapcomputer-science
user8657154
user8657154

Reputation:

Hash Tables and Separate Chaining: How do you know which value to return from the bucket's list?

We're learning about hash tables in my data structures and algorithms class, and I'm having trouble understanding separate chaining.

I know the basic premise: each bucket has a pointer to a Node that contains a key-value pair, and each Node contains a pointer to the next (potential) Node in the current bucket's mini linked list. This is mainly used to handle collisions.

Now, suppose for simplicity that the hash table has 5 buckets. Suppose I wrote the following lines of code in my main after creating an appropriate hash table instance.

myHashTable["rick"] = "Rick Sanchez";
myHashTable["morty"] = "Morty Smith";

Let's imagine whatever hashing function we're using just so happens to produce the same bucket index for both string keys rick and morty. Let's say that bucket index is index 0, for simplicity.

So at index 0 in our hash table, we have two nodes with values of Rick Sanchez and Morty Smith, in whatever order we decide to put them in (the first pointing to the second).

When I want to display the corresponding value for rick, which is Rick Sanchez per our code here, the hashing function will produce the bucket index of 0.

How do I decide which node needs to be returned? Do I loop through the nodes until I find the one whose key matches rick?

Upvotes: 7

Views: 2555

Answers (1)

Gabriel Avellaneda
Gabriel Avellaneda

Reputation: 742

To resolve Hash Tables conflicts, that's it, to put or get an item into the Hash Table whose hash value collides with another one, you will end up reducing a map to the data structure that is backing the hash table implementation; this is generally a linked list. In the case of a collision this is the worst case for the Hash Table structure and you will end up with an O(n) operation to get to the correct item in the linked list. That's it, a loop as you said, that will search the item with the matching key. But, in the cases that you have a data structure like a balanced tree to search, it can be O(logN) time, as the Java8 implementation.

As JEP 180: Handle Frequent HashMap Collisions with Balanced Trees says:

The principal idea is that once the number of items in a hash bucket grows beyond a certain threshold, that bucket will switch from using a linked list of entries to a balanced tree. In the case of high hash collisions, this will improve worst-case performance from O(n) to O(log n).

This technique has already been implemented in the latest version of the java.util.concurrent.ConcurrentHashMap class, which is also slated for inclusion in JDK 8 as part of JEP 155. Portions of that code will be re-used to implement the same idea in the HashMap and LinkedHashMap classes.

I strongly suggest to always look at some existing implementation. To say about one, you could look at the Java 7 implementation. That will increase your code reading skills, that is almost more important or you do more often than writing code. I know that it is more effort but it will pay off.

For example, take a look at the HashTable.get method from Java 7:

public synchronized V get(Object key) {
    Entry<?,?> tab[] = table;
    int hash = key.hashCode();
    int index = (hash & 0x7FFFFFFF) % tab.length;
    for (Entry<?,?> e = tab[index] ; e != null ; e = e.next) {
        if ((e.hash == hash) && e.key.equals(key)) {
            return (V)e.value;
        }
    }
    return null;
}

Here we see that if ((e.hash == hash) && e.key.equals(key)) is trying to find the correct item with the matching key.

And here is the full source code: HashTable.java

Upvotes: 2

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