706. Design HashMap

Problem

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 <= 10^6

• At most 10^4 calls will be made to put, get, and remove.

Solution (Java)

class MyHashMap {
    private ArrayList[] arr = null;

    public MyHashMap() {
        arr = new ArrayList[1000];
    }

    public void put(int key, int value) {
        int bucket = key % 1000;
        if (arr[bucket] == null) {
            ArrayList<Entry> list = new ArrayList<>();
            Entry e = new Entry();
            e.key = key;
            e.value = value;
            list.add(e);
            arr[bucket] = list;
        } else {
            ArrayList<Entry> list = arr[bucket];
            Entry e = new Entry();
            e.key = key;
            e.value = value;
            list.remove(e);
            list.add(e);
        }
    }

    public int get(int key) {
        int bucket = key % 1000;
        int ans = -1;
        ArrayList<Entry> list = arr[bucket];
        if (list != null) {
            for (Entry e : list) {
                if (e.key == key) {
                    ans = e.value;
                }
            }
        }
        return ans;
    }

    public void remove(int key) {
        int bucket = key % 1000;
        ArrayList<Entry> list = arr[bucket];
        Entry e = new Entry();
        e.key = key;
        if (list != null) {
            list.remove(e);
        }
    }

    static class Entry {
        int key;
        int value;

        @Override
        public int hashCode() {
            final int prime = 31;
            int result = 1;
            result = prime * result + key;
            return result;
        }

        @Override
        public boolean equals(Object obj) {
            if (this == obj) {
                return true;
            }
            if (obj == null) {
                return false;
            }
            if (getClass() != obj.getClass()) {
                return false;
            }
            Entry other = (Entry) obj;
            return key == other.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);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).