705. Design HashSet

Problem

Design a HashSet without using any built-in hash table libraries.

Implement MyHashSet class:

• void add(key) Inserts the value key into the HashSet.

• bool contains(key) Returns whether the value key exists in the HashSet or not.

• void remove(key) Removes the value key in the HashSet. If key does not exist in the HashSet, do nothing.

  Example 1:

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)

  Constraints:

• 0 <= key <= 10^6

• At most 10^4 calls will be made to add, remove, and contains.

Solution (Java)

class MyHashSet {
    private final boolean[] arr;

    public MyHashSet() {
        arr = new boolean[1000001];
    }

    public void add(int key) {
        arr[key] = true;
    }

    public void remove(int key) {
        arr[key] = false;
    }

    public boolean contains(int key) {
        return arr[key];
    }
}

/**
 * Your MyHashSet object will be instantiated and called as such:
 * MyHashSet obj = new MyHashSet();
 * obj.add(key);
 * obj.remove(key);
 * boolean param_3 = obj.contains(key);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).