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Design HashMap

Problem Statement​

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. Leetcode Link

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.

Code​

Python Code

class MyHashMap:
    def __init__(self):
        self.data = [None] * 1000001
    def put(self, key: int, val: int) -> None:
        self.data[key] = val
    def get(self, key: int) -> int:
        val = self.data[key]
        return val if val != None else -1
    def remove(self, key: int) -> None:
        self.data[key] = None