706. Design HashMap

Description

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

To be specific, your design should include these functions:

• put(key, value) : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.

• get(key): Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.

• remove(key) : Remove the mapping for the value key if this map contains the mapping for the key.

Constraints

• All keys and values will be in the range of [0, 1000000].

• The number of operations will be in the range of [1, 10000].

• Please do not use the built-in HashMap library.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

MyHashMap hashMap = new MyHashMap();

hashMap.put(1, 1);

hashMap.put(2, 2);

hashMap.get(1); // returns 1

hashMap.get(3); // returns -1 (not found)

hashMap.put(2, 1); // update the existing value

hashMap.get(2); // returns 1

hashMap.remove(2); // remove the mapping for 2

hashMap.get(2); // returns -1 (not found)

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class MyHashMap {
    private final Integer[] map;
    /** Initialize your data structure here. */
    public MyHashMap() {
        map = new Integer[1000000];
    }
    
    /** value will always be non-negative. */
    public void put(int key, int value) {
        map[key] = value;
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if 
        this map contains no mapping for the key */
    public int get(int key) {
        Integer value = map[key];
        return value == null? -1: value;
    }
    
    /** Removes the mapping of the specified value key if this map 
        contains a mapping for the key */
    public void remove(int key) {
        map[key] = null;
    }
}

/**
 * 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);
 */

Solution 2

/**
 * Time complexity : For each of the methods, the time complexity is O(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.
 * Space complexity : 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.
 */

class KVBlock {
    int key;
    int value;
    public KVBlock(int key, int value) {
        this.key = key;
        this.value = value;
    }
}

class KVChain {
    private int modKey;
    private List<KVBlock> kvChain;
    
    public KVChain(int modKey) {
        this.modKey = modKey;
        kvChain = new ArrayList<KVBlock>();
    }
    
    public int getValue(int key) {
        for(KVBlock kvBlock: kvChain) {
            if(kvBlock.key == key) {
                return kvBlock.value;
            }
        }
        return -1;
    }
    
    public void addValue(int key, int value) {
        for(KVBlock kvBlock: kvChain) {
            if(kvBlock.key == key) {
                kvBlock.value = value;
                return;
            }
        }
        kvChain.add(new KVBlock(key, value));
    }
    
    public void removeKey(int key) {
        for(int i = 0; i < kvChain.size(); i++) {
            if(kvChain.get(i).key == key) {
                kvChain.remove(i);
                return;
            }
        }
    }
}

class MyHashMap {
    
    private KVChain[] map;
    private int modKey = 2069;

    /** Initialize your data structure here. */
    public MyHashMap() {
        map = new KVChain[modKey];
    }
    
    /** value will always be non-negative. */
    public void put(int key, int value) {
        int index = key % modKey;
        if(map[index] == null) {
            map[index] = new KVChain(modKey);
        }
        map[index].addValue(key, value);
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if 
        this map contains no mapping for the key */
    public int get(int key) {
        int index = key % modKey;
        if(map[index] == null) {
            return -1;
        }
        return map[index].getValue(key);
    }
    
    /** Removes the mapping of the specified value key if this map contains 
        a mapping for the key */
    public void remove(int key) {
        int index = key % modKey;
        if(map[index] == null) {
            return;
        }
        map[index].removeKey(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);
 */

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