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Tested Python solution for LeetCode 705 with 13 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 705 # by problem number
lcpy gen -s design_hash_set # by problem name
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.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)
add, remove, and contains.class MyHashSet:
# Time: O(1)
# Space: O(n)
def __init__(self) -> None:
# Problem guarantees 0 <= key <= 10^6; a direct-address table is simplest.
self._present: list[bool] = [False] * 1_000_001
# Time: O(1)
# Space: O(1)
def add(self, key: int) -> None:
self._present[key] = True
# Time: O(1)
# Space: O(1)
def remove(self, key: int) -> None:
self._present[key] = False
# Time: O(1)
# Space: O(1)
def contains(self, key: int) -> bool:
return self._present[key]
| Time | Space |
|---|---|
| O(1) | O(n) |