# LeetCode 706. Design HashMap

## Description

https://leetcode.com/problems/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`.

## Explanation

A simple way is to create a big array. The indices of array are keys, values are corresponding values.

## Python Solution

``````class MyHashMap:

def __init__(self):
"""
"""
self.keys = [-1 for _ in range(1000000)]

def put(self, key: int, value: int) -> None:
"""
value will always be non-negative.
"""
if key >= len(self.keys):
for _ in range(1000000):
self.keys.append(-1)

self.keys[key] = value

def get(self, key: int) -> int:
"""
Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
"""
return self.keys[key]

def remove(self, key: int) -> None:
"""
Removes the mapping of the specified value key if this map contains a mapping for the key
"""
self.keys[key] = -1

# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)``````
• Time Complexity: O(N).
• Space Complexity: O(N).