Introduction
You already know the magic behind hash maps. Now it’s time to write your own implementation!
Limitation
Before we get started, we need to lay down some ground rules. JavaScript’s dynamic nature of arrays allows us to insert and retrieve indexes that are outside our array size range. Example: if we create an array of size 16
to represent our buckets, nothing stops us from storing items at index 500
. This defeats the purpose of limiting storage size in hash maps, so we need to enforce some restrictions.
Use the following snippet whenever you access a bucket through an index. We want to throw an error if we try to access an out-of-bounds index:
if (index < 0 || index >= buckets.length) {
throw new Error("Trying to access index out of bounds");
}
Assignment
Start by creating a HashMap
class or factory function. It’s up to you which you want to use. It should have at least two variables for load factor
and capacity
. Then proceed to create the following methods:
-
hash(key)
takes a key and produces a hash code with it. We already implemented a fairly goodhash
function in the previous lesson. As a reminder:function hash(key) { let hashCode = 0; const primeNumber = 31; for (let i = 0; i < key.length; i++) { hashCode = primeNumber * hashCode + key.charCodeAt(i); } return hashCode; }
You are free to use that, or you can conduct your own research on hashing algorithms. Beware, this is a deep, deep rabbit hole.
However, there is one edge case our
hash
function still needs to address. For very long keys, our hash code will exceed the maximum integer value allowed by JavaScript. Once that happens, calculations become inaccurate, and the chance of collisions significantly increases. One way to avoid this issue is to apply the modulo%
operator on each iteration instead of outside the loop at the end. This ensures the output never becomes larger than our bucket’s length.You might find yourself confusing keys with hash codes while accessing key-value pairs later. We would like to stress that the key is what your
hash
function will take as an input. In a way, we could say that the key is important for us only inside thehash
function, as we never access a bucket directly with the key. Instead, we always do so with the hash code.In the real world, hash maps can accommodate various data types as keys, including numbers, strings, or objects. However, for this project, we will only handle keys of type
string
. -
set(key, value)
takes two arguments: the first is a key, and the second is a value that is assigned to this key. If a key already exists, then the old value is overwritten, and we can say that we update the key’s value (e.g.Carlos
is our key but it is called twice: once with valueI am the old value.
, and once with valueI am the new value.
. Following this logic,Carlos
should contain only the latter value).Recall that collisions occur when TWO DIFFERENT keys generate the same hash code and get assigned to the same bucket. (e.g.
Carlos
andCarla
are both hashed to3
, so3
becomes a location forCarlos
ANDCarla
. However, we know that this is not an update because the keys are different). Review the dealing with collisions section of the previous lesson to find a way to handle our collisions.- Remember to grow your buckets to double their capacity when your hash map reaches the
load factor
. The methods mentioned later in this assignment can help you handle the growth logic, so you may want to implement this feature near the end. However, we mention this withset()
because it’s important to grow buckets exactly as they are being expanded.
- Remember to grow your buckets to double their capacity when your hash map reaches the
-
get(key)
takes one argument as a key and returns the value that is assigned to this key. If a key is not found, returnnull
. -
has(key)
takes a key as an argument and returnstrue
orfalse
based on whether or not the key is in the hash map. -
remove(key)
takes a key as an argument. If the given key is in the hash map, it should remove the entry with that key and returntrue
. If the key isn’t in the hash map, it should returnfalse
. -
length()
returns the number of stored keys in the hash map. -
clear()
removes all entries in the hash map. -
keys()
returns an array containing all the keys inside the hash map. -
values()
returns an array containing all the values. -
entries()
returns an array that contains eachkey, value
pair. Example:[[firstKey, firstValue], [secondKey, secondValue]]
Remember that a hash map does not preserve insertion order when you are retrieving your hash map’s data. It is normal and expected for keys and values to appear out of the order you inserted them in.
Test Your Hash Map
-
Create a new JavaScript file.
-
Create a new instance of your hash map and set the load factor to be
0.75
.const test = new HashMap() // or HashMap() if using a factory
-
Populate your hash map using the
set(key, value)
method by copying the following:test.set('apple', 'red') test.set('banana', 'yellow') test.set('carrot', 'orange') test.set('dog', 'brown') test.set('elephant', 'gray') test.set('frog', 'green') test.set('grape', 'purple') test.set('hat', 'black') test.set('ice cream', 'white') test.set('jacket', 'blue') test.set('kite', 'pink') test.set('lion', 'golden')
-
After populating your hash map with the data above, your hash map’s current load levels should now be at
0.75
(full capacity). -
Now with a full hash map, try overwriting a few nodes using
set(key, value)
. This should only overwrite the existingvalues
of your nodes and not add new ones, solength()
should still return the same value andcapacity
should remain the same. -
After that, populate your hash map with the last node below. This will make your load levels exceed your
load factor
, triggering your hash map’s growth functionality and doubling itscapacity
:test.set('moon', 'silver')
-
If you have implemented your hash map correctly, the load levels of your expanded hash map should drop well below your load factor, and the entries should be spread evenly among the expanded buckets.
-
With your new hash map, try overwriting a few nodes using
set(key, value)
. Again, this should only overwrite existingvalues
of your nodes. -
Test the other methods of your hash map, such as
get(key)
,has(key)
,remove(key)
,length()
,clear()
,keys()
,values()
, andentries()
, to check if they are still working as expected after expanding your hash map.
Extra Credit
- Create a
HashSet
class or factory function that behaves the same as aHashMap
but only containskeys
with novalues
.