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

My code

class MyHashMap:

    def __init__(self):
        self.map: dict = {}

    def put(self, key: int, value: int) -> None:
        self.map[key] = value

    def get(self, key: int) -> int:
        return self.map.get(key, -1)
        

    def remove(self, key: int) -> None:
        if self.map.get(key):
            del self.map[key]

Review

[실행 결과]
Runtime 299 ms
Memory Usage: 17.1 MB

[접근법]
다른 분들 푸신 방법 보고 dict로 풀면 짧게 끝낼 수 있겠다 해서 여기로 접근했다.
get에서 mapped된 key가 있으면 리턴하고 아니면 -1리턴
remove에서 맵이 key값을 갖고 있으면 해당 값을 지운다

[느낀점]
혼자 떠올릴 수 있었으면 좋겠다.