How do you solve Design HashMap on LeetCode?

Design HashMap (LeetCode #706) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) average case time complexity and O(n) space complexity. Key insight: Understanding hash functions is crucial to efficiently mapping keys to indices.

What is the brute force solution for Design HashMap?

The brute force approach for Design HashMap is Brute Force, with O(n²) time complexity and O(1) space complexity. In a brute-force approach, we can use a simple list to store key-value pairs. Each time we want to insert or retrieve a value, we will search through the list, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1) average case.

What algorithm or data structure is used to solve Design HashMap?

Design HashMap uses the following concepts: Array, Hash Table, Linked List, Design, Hash Function. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design HashMap?

Explain your thought process clearly while coding. Discuss potential edge cases and how you would handle them. Practice implementing hash tables with different collision resolution strategies.

What are common mistakes when solving Design HashMap?

Forgetting to handle collisions when multiple keys hash to the same index. Not initializing the linked list at each index before adding key-value pairs.

#706

Design HashMap

Easy

ArrayHash TableLinked ListDesignHash FunctionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1) average case
Space	O(1)	O(n)

💡

Intuition

Time O(1) average caseSpace O(n)

An optimal solution uses a hash table with an array of linked lists (or buckets) to store key-value pairs. This allows for efficient insertion, retrieval, and deletion operations by minimizing collisions.

⚙️

Algorithm

5 steps

1Step 1: Initialize an array of linked lists to store key-value pairs.
2Step 2: Use a hash function to compute the index for the key.
3Step 3: For 'put', check if the key exists in the linked list at the computed index; if it does, update the value; if not, add a new node.
4Step 4: For 'get', compute the index and search the linked list for the key, returning the corresponding value or -1 if not found.
5Step 5: For 'remove', compute the index and search the linked list to find and remove the node with the key.

solution.py51 lines

1class ListNode:
2    def __init__(self, key, value):
3        self.key = key
4        self.value = value
5        self.next = None
6
7class MyHashMap:
8    def __init__(self):
9        self.size = 1000
10        self.map = [None] * self.size
11
12    def _hash(self, key):
13        return key % self.size
14
15    def put(self, key: int, value: int) -> None:
16        index = self._hash(key)
17        if not self.map[index]:
18            self.map[index] = ListNode(-1, -1)
19        prev = self.map[index]
20        curr = prev.next
21        while curr:
22            if curr.key == key:
23                curr.value = value
24                return
25            prev = curr
26            curr = curr.next
27        prev.next = ListNode(key, value)
28
29    def get(self, key: int) -> int:
30        index = self._hash(key)
31        if not self.map[index]:
32            return -1
33        curr = self.map[index].next
34        while curr:
35            if curr.key == key:
36                return curr.value
37            curr = curr.next
38        return -1
39
40    def remove(self, key: int) -> None:
41        index = self._hash(key)
42        if not self.map[index]:
43            return
44        prev = self.map[index]
45        curr = prev.next
46        while curr:
47            if curr.key == key:
48                prev.next = curr.next
49                return
50            prev = curr
51            curr = curr.next

ℹ

Complexity note: The average time complexity for put, get, and remove operations is O(1) due to the efficient use of hash functions and linked lists to handle collisions. However, in the worst case (if many keys hash to the same index), it can degrade to O(n).

1Understanding hash functions is crucial to efficiently mapping keys to indices.
2Handling collisions properly is essential for maintaining performance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.