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. For a load factor of 0.75 you should have an initial capacity of size 16. Then proceed to create the following methods:

1. hash(key) takes a string key and produces a hash code with it (in the real world, hash maps can accommodate various data types as keys, such as numbers or even other objects, but we’ll keep it simple for now). We already implemented a fairly good hash 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. Remember that to ensure we get an index that fits in our buckets (regardless of growth), we need to modulo (%) the hash code by our current capacity before we return it. An important edge case to note is that with very long keys, we may end up exceeding the maximum safe integer and calculations end up inaccurate. Therefore, it’d be sensible to modulo the hash code by the current capacity on every iteration in the loop, rather than only once after the loop has completed.

   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 the hash function, as we never access a bucket directly with the key. Instead, we always do so with the hash code.

2. 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 value I am the old value., and once with value I 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. Rama and Sita are both hashed to 3, so 3 becomes a location for Rama AND Sita. 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 with set() because it’s important to grow buckets exactly as they are being expanded.

3. get(key) takes one argument as a key and returns the value that is assigned to this key. If a key is not found, return null.

4. has(key) takes a key as an argument and returns true or false based on whether or not the key is in the hash map.

5. 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 return true. If the key isn’t in the hash map, it should return false.

6. length() returns the number of stored keys in the hash map.

7. clear() removes all entries in the hash map.

8. keys() returns an array containing all the keys inside the hash map.

9. values() returns an array containing all the values.

10. entries() returns an array that contains each key, 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

1. Create a new JavaScript file.

2. 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
   

3. 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')
   

4. After populating your hash map with the data above, your hash map’s current load levels should now be at 0.75 (full capacity).

5. Now with a full hash map, try overwriting a few nodes using set(key, value). This should only overwrite the existing values of your nodes and not add new ones, so length() should still return the same value and capacity should remain the same.

6. 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 its capacity:

   test.set('moon', 'silver')
   

7. 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.

8. With your new hash map, try overwriting a few nodes using set(key, value). Again, this should only overwrite existing values of your nodes.

9. Test the other methods of your hash map, such as get(key), has(key), remove(key), length(), clear(), keys(), values(), and entries(), 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 a HashMap but only contains keys with no values.