Design HashMap

Design a HashMap without using any built-in hash table libraries.

Implement the MyHashMap class:

• MyHashMap() initializes the object with an empty map.
• void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
• int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
• void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.

 

Example 1:

Input
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]

Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1);    // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3);    // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2);    // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2);    // return -1 (i.e., not found), The map is now [[1,1]]

 

Constraints:

• 0 <= key, value <= 106
• At most 104 calls will be made to put, get, and remove.

class Pair<U,V>{
    public U first;
    public V second;
    public Pair(U first, V second){
        this.first = first;
        this.second = second;
    }
}

class Bucket{
    private List<Pair<Integer,Integer>> bucket;
    public Bucket(){
        this.bucket = new LinkedList<Pair<Integer,Integer>>();
    }
    
    public Integer get(Integer key){
        for(Pair<Integer,Integer> pair:this.bucket){
            if(pair.first.equals(key)){
                return pair.second;
            }
        }
        return -1;
    }
    
    public void update(Integer key,Integer value){
        boolean isFound = false;
        for(Pair<Integer,Integer> pair:this.bucket){
            if(pair.first.equals(key)){
                isFound = true;
                pair.second = value;
            }
        }
        if(!isFound){
            this.bucket.add(new Pair<Integer,Integer>(key,value));
        }
    }
    
    public void remove(Integer key){
        for(Pair<Integer,Integer> pair : this.bucket){
            if(pair.first.equals(key)){
                this.bucket.remove(pair);
                break;
            }
        }
    }
}



class MyHashMap {
    private int key_space;
    private List<Bucket> hash_table;

    public MyHashMap() {
        this.key_space = 2069;
        this.hash_table = new ArrayList<Bucket>();
        for(int i =0;i<key_space;i++){
            this.hash_table.add(new Bucket());
        }
    }
    
    public void put(int key, int value) {
        int hash_key = key % this.key_space;
        this.hash_table.get(hash_key).update(key,value);
    }
    
    public int get(int key) {
        int hash_key = key % this.key_space;
        return this.hash_table.get(hash_key).get(key);
    }
    
    public void remove(int key) {
        int hash_key = key % this.key_space;
        this.hash_table.get(hash_key).remove(key);
    }
}

/**
 * Your MyHashMap object will be instantiated and called as such:
 * MyHashMap obj = new MyHashMap();
 * obj.put(key,value);
 * int param_2 = obj.get(key);
 * obj.remove(key);
 */

Complexity Analysis
Time Complexity: for each of the methods, the time complexity is $\mathcal{O}(\frac{N}{K})$ where $N$ is the number of all possible keys and $K$ is the number of predefined buckets in the hashmap, which is 2069 in our case.
In the ideal case, the keys are evenly distributed in all buckets. As a result, on average, we could consider the size of the bucket is $\frac{N}{K}$.
Since in the worst case we need to iterate through a bucket to find the desire value, the time complexity of each method is $\mathcal{O}(\frac{N}{K})$.
Space Complexity: $\mathcal{O}(K+M)$ where $K$ is the number of predefined buckets in the hashmap and $M$ is the number of unique keys that have been inserted into the hashmap.