How do you solve Design HashSet on LeetCode?

Design HashSet (LeetCode #705) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) on average time complexity and O(n) space complexity. Key insight: Using a hash function allows for efficient indexing of elements.

What is the brute force solution for Design HashSet?

The brute force approach for Design HashSet is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can use a simple list to store the elements of the HashSet. For each operation, we will iterate through the list to check for existence or to remove an element. This is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(1) on average.

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

Design HashSet 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 HashSet?

Explain your thought process clearly during the interview. Discuss trade-offs between different data structures. Practice implementing hash functions and collision resolution techniques.

What are common mistakes when solving Design HashSet?

Not handling collisions properly, leading to degraded performance. Forgetting to check if an element exists before adding it.

#705

Design HashSet

Easy

ArrayHash TableLinked ListDesignHash FunctionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(1) on averageSpace O(n)

The optimal solution uses a combination of an array and linked lists to handle collisions, which allows for efficient storage and retrieval of elements. This approach ensures that both the average time complexity for operations is reduced significantly.

⚙️

Algorithm

5 steps

1Step 1: Create an array of buckets (lists) to store elements.
2Step 2: Use a hash function to determine the index for each key based on its value.
3Step 3: For 'add(key)', compute the index and add the key to the corresponding bucket if not present.
4Step 4: For 'contains(key)', compute the index and check if the key exists in the corresponding bucket.
5Step 5: For 'remove(key)', compute the index and remove the key from the corresponding bucket if it exists.

solution.py21 lines

1class MyHashSet:
2    def __init__(self):
3        self.size = 1000
4        self.buckets = [[] for _ in range(self.size)]
5
6    def _hash(self, key):
7        return key % self.size
8
9    def add(self, key):
10        bucket_index = self._hash(key)
11        if key not in self.buckets[bucket_index]:
12            self.buckets[bucket_index].append(key)
13
14    def contains(self, key):
15        bucket_index = self._hash(key)
16        return key in self.buckets[bucket_index]
17
18    def remove(self, key):
19        bucket_index = self._hash(key)
20        if key in self.buckets[bucket_index]:
21            self.buckets[bucket_index].remove(key)

ℹ

Complexity note: The average time complexity for add, contains, and remove operations is O(1) due to the use of a hash function that distributes keys uniformly across the buckets. However, in the worst case (e.g., many collisions), it can degrade to O(n). The space complexity is O(n) because we store each key in the buckets.

1Using a hash function allows for efficient indexing of elements.
2Collision handling is crucial for maintaining performance.

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