Introduction

In this blog post, we will explore how to solve the LeetCode problem "706" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. 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

Here's a solution to the HashMap implementation problem, focusing on efficiency and clarity:

• Chaining with a Fixed-Size Array: Use a fixed-size array as the underlying storage. Each index in the array represents a "bucket." Collisions (when different keys map to the same index) are handled using separate chaining. Each bucket stores a linked list of key-value pairs.
• Simple Hash Function: Implement a simple hash function (e.g., modulo operation) to map keys to array indices. The goal is to distribute keys relatively evenly across the buckets to minimize collisions.

• Linked List Operations: put, get, and remove operations involve traversing the linked list at the bucket corresponding to the key's hash.

• Runtime Complexity: O(1) average case for put, get, and remove, assuming a good hash function and relatively uniform key distribution. In the worst-case scenario (all keys hash to the same bucket), the complexity degrades to O(n), where n is the number of keys. Storage Complexity: O(n), where n is the number of key-value pairs stored in the HashMap.

Code

    class MyHashMap:

    def __init__(self):
        self.capacity = 1000  # Choose a reasonable capacity
        self.table = [None] * self.capacity

    def _hash(self, key):
        return key % self.capacity

    def put(self, key: int, value: int) -> None:
        index = self._hash(key)
        if self.table[index] is None:
            self.table[index] = [(key, value)]
        else:
            for i, (k, v) in enumerate(self.table[index]):
                if k == key:
                    self.table[index][i] = (key, value)
                    return
            self.table[index].append((key, value))

    def get(self, key: int) -> int:
        index = self._hash(key)
        if self.table[index] is None:
            return -1
        else:
            for k, v in self.table[index]:
                if k == key:
                    return v
            return -1

    def remove(self, key: int) -> None:
        index = self._hash(key)
        if self.table[index] is not None:
            for i, (k, v) in enumerate(self.table[index]):
                if k == key:
                    del self.table[index][i]
                    return


# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)