Description
https://leetcode.com/problems/design-hashset/
Design a HashSet without using any built-in hash table libraries.
Implement MyHashSet
class:
void add(key)
Inserts the valuekey
into the HashSet.bool contains(key)
Returns whether the valuekey
exists in the HashSet or not.void remove(key)
Removes the valuekey
in the HashSet. Ifkey
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 <= 106
- At most
104
calls will be made toadd
,remove
, andcontains
.
Follow up: Could you solve the problem without using the built-in HashSet library?
Explanation
In Python, we can implement with a list.
Python Solution
class MyHashSet:
def __init__(self):
"""
Initialize your data structure here.
"""
self.hash_set = []
def add(self, key: int) -> None:
if key not in self.hash_set:
self.hash_set.append(key)
def remove(self, key: int) -> None:
if key in self.hash_set:
self.hash_set.remove(key)
def contains(self, key: int) -> bool:
"""
Returns true if this set contains the specified element
"""
return key in self.hash_set
# Your MyHashSet object will be instantiated and called as such:
# obj = MyHashSet()
# obj.add(key)
# obj.remove(key)
# param_3 = obj.contains(key)
- Time Complexity: O(N).
- Space Complexity: O(N).