How does a Java HashMap handle different objects with the same hash code?

Question

As per my understanding I think:

1. It is perfectly legal for two objects to have the same hashcode.
2. If two objects are equal (using the equals() method) then they have the same hashcode.
3. If two objects are not equal then they cannot have the same hashcode

Am I correct?

Now if am correct, I have the following question: The HashMap internally uses the hashcode of the object. So if two objects can have the same hashcode, then how can the HashMap track which key it uses?

Can someone explain how the HashMap internally uses the hashcode of the object?

Solution 1

A hashmap works like this (this is a little bit simplified, but it illustrates the basic mechanism):

It has a number of "buckets" which it uses to store key-value pairs in. Each bucket has a unique number - that's what identifies the bucket. When you put a key-value pair into the map, the hashmap will look at the hash code of the key, and store the pair in the bucket of which the identifier is the hash code of the key. For example: The hash code of the key is 235 -> the pair is stored in bucket number 235. (Note that one bucket can store more then one key-value pair).

When you lookup a value in the hashmap, by giving it a key, it will first look at the hash code of the key that you gave. The hashmap will then look into the corresponding bucket, and then it will compare the key that you gave with the keys of all pairs in the bucket, by comparing them with equals().

Now you can see how this is very efficient for looking up key-value pairs in a map: by the hash code of the key the hashmap immediately knows in which bucket to look, so that it only has to test against what's in that bucket.

Looking at the above mechanism, you can also see what requirements are necessary on the hashCode() and equals() methods of keys:

• If two keys are the same (equals() returns true when you compare them), their hashCode() method must return the same number. If keys violate this, then keys that are equal might be stored in different buckets, and the hashmap would not be able to find key-value pairs (because it's going to look in the same bucket).

• If two keys are different, then it doesn't matter if their hash codes are the same or not. They will be stored in the same bucket if their hash codes are the same, and in this case, the hashmap will use equals() to tell them apart.

Solution 2

Your third assertion is incorrect.

It's perfectly legal for two unequal objects to have the same hash code. It's used by HashMap as a "first pass filter" so that the map can quickly find possible entries with the specified key. The keys with the same hash code are then tested for equality with the specified key.

You wouldn't want a requirement that two unequal objects couldn't have the same hash code, as otherwise that would limit you to 2³² possible objects. (It would also mean that different types couldn't even use an object's fields to generate hash codes, as other classes could generate the same hash.)

Solution 3

HashMap is an array of Entry objects.

Consider HashMap as just an array of objects.

Have a look at what this Object is:

static class Entry<K,V> implements Map.Entry<K,V> {
        final K key;
        V value;
        Entry<K,V> next;
        final int hash;
… 
}

Each Entry object represents a key-value pair. The field next refers to another Entry object if a bucket has more than one Entry.

Sometimes it might happen that hash codes for 2 different objects are the same. In this case, two objects will be saved in one bucket and will be presented as a linked list. The entry point is the more recently added object. This object refers to another object with the next field and so on. The last entry refers to null.

When you create a HashMap with the default constructor

HashMap hashMap = new HashMap();

The array is created with size 16 and default 0.75 load balance.

Adding a new key-value pair

1. Calculate hashcode for the key
2. Calculate position hash % (arrayLength-1) where element should be placed (bucket number)
3. If you try to add a value with a key which has already been saved in HashMap, then value gets overwritten.
4. Otherwise element is added to the bucket.

If the bucket already has at least one element, a new one gets added and placed in the first position of the bucket. Its next field refers to the old element.

Deletion

1. Calculate hashcode for the given key
2. Calculate bucket number hash % (arrayLength-1)
3. Get a reference to the first Entry object in the bucket and by means of equals method iterate over all entries in the given bucket. Eventually we will find the correct Entry. If a desired element is not found, return null

Solution 4

You can find excellent information at http://javarevisited.blogspot.com/2011/02/how-hashmap-works-in-java.html

To Summarize:

HashMap works on the principle of hashing

put(key, value): HashMap stores both key and value object as Map.Entry. Hashmap applies hashcode(key) to get the bucket. if there is collision ,HashMap uses LinkedList to store object.

get(key): HashMap uses Key Object's hashcode to find out bucket location and then call keys.equals() method to identify correct node in LinkedList and return associated value object for that key in Java HashMap.

Solution 5

Here is a rough description of HashMap's mechanism, for Java 8 version, (it might be slightly different from Java 6).

---

Data structures

• Hash table
  Hash value is calculated via hash() on key, and it decide which bucket of the hashtable to use for a given key.
• Linked list (singly)
  When count of elements in a bucket is small, a singly linked list is used.
• Red-Black tree
  When count of elements in a bucket is large, a red-black tree is used.

---

Classes (internal)

• Map.Entry
  Represent a single entity in map, the key/value entity.

• HashMap.Node
  Linked list version of node.

  It could represent:

  • A hash bucket.
    Because it has a hash property.
  • A node in singly linked list, (thus also head of linkedlist).
• HashMap.TreeNode
  Tree version of node.

---

Fields (internal)

• Node[] table
  The bucket table, (head of the linked lists).
  If a bucket don't contains elements, then it's null, thus only take space of a reference.
• Set<Map.Entry> entrySet Set of entities.
• int size
  Number of entities.
• float loadFactor
  Indicate how full the hash table is allowed, before resizing.
• int threshold
  The next size at which to resize.
  Formula: threshold = capacity * loadFactor

---

Methods (internal)

• int hash(key)
  Calculate hash by key.

• How to map hash to bucket?
  Use following logic:

  static int hashToBucket(int tableSize, int hash) {
      return (tableSize - 1) & hash;
  }
  

---

About capacity

In hash table, capacity means the bucket count, it could be get from table.length.
Also could be calculated via threshold and loadFactor, thus no need to be defined as a class field.

Could get the effective capacity via: capacity()

---

Operations

• Find entity by key.
  First find the bucket by hash value, then loop linked list or search sorted tree.
• Add entity with key.
  First find the bucket according to hash value of key.
  Then try find the value:
  • If found, replace the value.
  • Otherwise, add a new node at beginning of linked list, or insert into sorted tree.
• Resize
  When threshold reached, will double hashtable's capacity(table.length), then perform a re-hash on all elements to rebuild the table.
  This could be an expensive operation.

---

Performance

• get & put
  Time complexity is O(1), because:
  • Bucket is accessed via array index, thus O(1).
  • Linked list in each bucket is of small length, thus could view as O(1).
  • Tree size is also limited, because will extend capacity & re-hash when element count increase, so could view it as O(1), not O(log N).

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Comments

[Joachim Sauer]

For the record: #1 and #2 are correct, #3 is wrong: Two objects that are not equal may have the same hash code.

[Delfic]

#1 and #3 are contradictory even

[Joachim Sauer]

Indeed, if #2 is not followed, then the equals() implementation (or arguably the hashCode()) is incorrect.

[Geek]

how did u arrive at 2^32 possible objects ?

[Geek]

you wrote "and the hashmap would not be able to find key-value pairs (because it's going to look in the same bucket)." Can you explain it is going to look in the same bucket say those two equals objects are t1 and t2 and they are equal and t1 and t2 have hashcodes h1 and h2 respectively .So t1.equals(t2)=true and h1!=h2 So when the hashmap would look for t1 , it will look in the bucket h1 and for t2 in the bucket t2 ?

[Jesper]

If two keys are equal but their hashCode() method returns different hash codes, then the equals() and hashCode() methods of the key class violate the contract and you'll get strange results when using those keys in a HashMap.

[Ankit Sharma]

Each bucket can have multiple Key Values pairs, which are uses linked list internally. But my confusion is - what is bucket here? What data structure it uses internally? Is there any connection between buckets?

[Jesper]

@AnkitSharma If you want to really know all the details, lookup the source code of HashMap, which you can find in the file src.zip in your JDK installation directory.

[Ankit Sharma]

Thanks @Jesper I got the solution - Data structure to store Entry objects is an array named table of type Entry. A particular index location in array is referred as bucket, because it can hold the first element of a LinkedList of Entry objects.

[Ankit Sharma]

Refer this link for more details howtodoinjava.com/2012/10/09/how-hashmap-works-in-java

[Narendra N]

I found the answer provided by Jasper better, I felt the blog is more towards handling interview, than understanding the concept

[Abhijit Gaikwad]

@NarendraN I agree with you.

[LittleLittleQ]

But for efficiency, you'd better keep your different object having different hashcode

[Jesper]

@AnnabellChan Yes, indeed. A HashMap would still work if all keys had the same hashcode, but it would be very inefficient, because everything would go into one bucket. Finding something in the map would then be just as inefficient as finding it in a list (O(n)).

[Admin]

@Jesper Really great explanation, helped me understand it very well. I know this might be a little late but when I was reading your answer I had a quick question. In the buckets of the HashMap which hold the key/value pairs, is there any relationship amongst the keys in the same bucket? Like are the keys within a specific range or something like that? I read online there is some relation but not too sure how it actually determines this relationship between the keys in a particular bucket...

[Jesper]

@1290 The only relation between the keys in the same bucket is that they have the same hash code.

[Admin]

@Jesper Oh ok, Thank you for clarifying.

[Navrattan Yadav]

why it use bucket size power of 2 ?

[ACV]

I think this is wrong: "It stores keys in an array and values in a LinkedList ."

[ACV]

each element in the list for each bucket contains the key and the value as well as the reference to the next node.

[weston]

This is wrong hash % (arrayLength-1) it would be hash % arrayLength. But it actually is hash & (arrayLength-1). That is, because it uses powers of two (2^n) for array length, taking n least significant bits.

[weston]

@navy It's a good growing strategy to double up. But other bits of the code rely on it, check out indexFor where it works out the bucket to use.

[Mudit bhaintwal]

I think it's not an array of Entity objects rather an array of LinkedList/Tree. And each tree internally have Entity objects.

[Jitendra]

Can you please give a example How has time complexity O(1)

[Eric]

@jsroyal This might explain the complexity more clearly: en.wikipedia.org/wiki/Hash_table. But in short: finding the target bucket is O(1), because you find it by index in an array; then within a bucket, the amount is elements is small & on average a constant number despite the total number of elements in the whole hashtable, so searching for the target element within a bucket is also O(1); thus, O(1) + O(1) = O(1).

[kiedysktos]

@Jesper thanks a lot, this is one of the best explanations on this topic I've seen

[roottraveller]

@shevchyk why do we store key and hash? what are their use ? Are not we wasting memory here?

[Sudhanshu Gaur]

awesome answer (y)

[overexchange]

hashset internally uses hashmap. does addition and deletion rules of hashmap, holds good for hashset?

[xeruf]

I'm late, but for those still wondering: A hashcode in Java is an int, and an int has 2^32 possible values

[Eugene]

@weston not only that, hashCode is an int which of course can be negative, doing modulo on a negative number will give you a negative number

[Sharan Rai]

This answer need to be edited, once after the hashcode is calculated, bucket number in which "key, value,hashcode info" will be stored need to be calculated, because its not a good idea to have as many numbers of buckets in as per hashcode number. for Example 235%16(initial capacity in java) 11 is the bucket number in above case.