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Leetcode #705: Design HashSet

In this guide, we solve Leetcode #705 Design HashSet in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Design a HashSet without using any built-in hash table libraries. Implement MyHashSet class: void add(key) Inserts the value key into the HashSet.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Design, Array, Hash Table, Linked List, Hash Function

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

Input ["MyHashSet", "add", "add", "contains", "contains", "add", "contains", "remove", "contains"] [[], [1], [2], [1], [3], [2], [2], [2], [2]] Output [null, null, null, true, false, null, true, null, false] Explanation MyHashSet myHashSet = new MyHashSet(); myHashSet.add(1); // set = [1] myHashSet.add(2); // set = [1, 2] myHashSet.contains(1); // return True myHashSet.contains(3); // return False, (not found) myHashSet.add(2); // set = [1, 2] myHashSet.contains(2); // return True myHashSet.remove(2); // set = [1] myHashSet.contains(2); // return False, (already removed)

Python Solution

class MyHashSet: def __init__(self): self.data = [False] * 1000001 def add(self, key: int) -> None: self.data[key] = True def remove(self, key: int) -> None: self.data[key] = False def contains(self, key: int) -> bool: return self.data[key] # Your MyHashSet object will be instantiated and called as such: # obj = MyHashSet() # obj.add(key) # obj.remove(key) # param_3 = obj.contains(key)

Complexity

The time complexity is O(n). The space complexity is O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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