Design a hashmap without using any built-in libraries

Design a hashmap without using any built-in libraries. You should include the following functions

• put(key, value): Insert a (key, value) pair into the hashmap. If the value already exists, update the value.
• get(key): Returns the value to which key is mapped, or -1 if this map contains nothing for key.
• remove(key): Remove the mapping for the value in key if it exists.

With JS Objects

I think the most direct way of implementing this would be using an JS Object, since it’s essentially what we need for direct access, with O(1) and all good!.

class ObjectHashMap<T> {
  constructor(protected readonly data: Record<string, T> = {}) {}

  get(key: string): T | -1 {
    return this.data[key] ?? -1
  }

  put(key: string, value: T): void {
    this.data[key] = value
  }

  remove(key: string): void {
    delete this.data[key]
  }
}

---

With Arrays (2D-Array)

Cassidy will say that using objects is cheating tho, so we should probably try to achieve the same with arrays!. Good luck getting the same complexity :(

interface MapItem<T> {
  readonly key: string
  readonly value: T
}

/** HashMap implementation using a 2D array */
class ArrayHashMap<T> {
  /**
   * Buckets where data will be stored
   *
   * <key hash % max> => [ <item1>, <item2>, ..., <itemN> ]
   */
  protected readonly buckets: MapItem<T>[][] = []

  /**
   * Constructor.
   * @param bucketSize Size of the array, greater means less collisions.
   */
  constructor(protected readonly bucketSize: number = 1000) {}

  /**
   * Gets a value from the map.
   * @param key
   * @returns value if found, -1 otherwise.
   */
  get(key: string): T | -1 {
    const bucket = this.getBucket(key)
    const item = bucket.find((item) => item.key === key)

    return item !== undefined ? item.value : -1
  }

  /**
   * Adds a new value to the map, replaces it if it existed.
   * @param key
   * @param value
   */
  put(key: string, value: T): void {
    const bucket = this.getBucket(key)
    for (const item of bucket) {
      if (item.key === key) {
        item.value = value
        return
      }
    }

    const index = this.getBucketIndex(key)
    bucket.push({ key, value })
    this.buckets[index] = bucket
  }

  /**
   * Removes a key from the map.
   * @param key
   */
  remove(key: string): void {
    const index = this.getBucketIndex(key)
    const updatedBucket = this.getBucket(key).filter((item) => item.key !== key)

    this.buckets[index] = updatedBucket
  }

  /**
   * Gets the bucket where the key should be present.
   * @param key
   * @returns bucket or empty array if the key is not present.
   */
  protected getBucket(key: string): MapItem<T>[] {
    const index = this.getHash(key) % this.bucketSize

    return this.buckets[index] || []
  }

  /**
   * Gets the bucket index from a key, ensures the index is not outside the
   * boundaries of the array. (ha, as if that mattered at all here).
   * @param key
   * @returns index
   */
  protected getBucketIndex(key: string): number {
    return this.getHash(key) % this.bucketSize
  }

  /**
   * Gets a hash from a string,
   * unsurprisingly, I took this from Stack-overflow (:
   * @param key
   * @returns numeric hash from a string
   */
  protected getHash(key: string): number {
    return key.split('').reduce((result: number, char: string) => {
      result = (result << 5) - result + char.charCodeAt(0)

      return result & result
    }, 0)
  }
}

Actually, this solution is a bit of a lie, since arrays are objects behind the scenes in JS, but okay… This solution is prepared for other languages where we have static arrays that allocate a specific amount of memory (bucketSize). The more array positions we reserve, the less probable collisions become (thus we get better performance). Complexity for this one goes from O(1) when the item is the only element in a bucket, to O(n) in the worst case where there is only one bucket and all the elements are placed in that one (oh boy, if that is not bad luck!).

Using JS here feels a bit like cheating though, since the first array and the array inside the buckets are the same kind of array (that are actually objects), and the bucketSize with the % operator is futile, since any “position” of the array is alway accessible without reserving memory for all the positions before. It would be fun to implement this using Rust, for example, where I would need to use a real array for the buckets, and a linked list (or whatever) for the list inside the buckets.