Leetcode #1206: Design Skiplist
In this guide, we solve Leetcode #1206 Design Skiplist 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.
Problem Statement
Design a Skiplist without using any built-in libraries. A skiplist is a data structure that takes O(log(n)) time to add, erase and search.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Linked List
Intuition
Linked list problems often require pointer manipulation rather than extra memory.
Two-pointer techniques expose cycles, midpoints, or reordering patterns.
Approach
Traverse with fast/slow pointers or reverse sublists when needed.
Maintain invariants carefully to avoid losing nodes.
Steps:
- Traverse with pointers.
- Reverse or split if required.
- Reconnect nodes correctly.
Example
Input
["Skiplist", "add", "add", "add", "search", "add", "search", "erase", "erase", "search"]
[[], [1], [2], [3], [0], [4], [1], [0], [1], [1]]
Output
[null, null, null, null, false, null, true, false, true, false]
Explanation
Skiplist skiplist = new Skiplist();
skiplist.add(1);
skiplist.add(2);
skiplist.add(3);
skiplist.search(0); // return False
skiplist.add(4);
skiplist.search(1); // return True
skiplist.erase(0); // return False, 0 is not in skiplist.
skiplist.erase(1); // return True
skiplist.search(1); // return False, 1 has already been erased.
Python Solution
class Node:
__slots__ = ['val', 'next']
def __init__(self, val: int, level: int):
self.val = val
self.next = [None] * level
class Skiplist:
max_level = 32
p = 0.25
def __init__(self):
self.head = Node(-1, self.max_level)
self.level = 0
def search(self, target: int) -> bool:
curr = self.head
for i in range(self.level - 1, -1, -1):
curr = self.find_closest(curr, i, target)
if curr.next[i] and curr.next[i].val == target:
return True
return False
def add(self, num: int) -> None:
curr = self.head
level = self.random_level()
node = Node(num, level)
self.level = max(self.level, level)
for i in range(self.level - 1, -1, -1):
curr = self.find_closest(curr, i, num)
if i < level:
node.next[i] = curr.next[i]
curr.next[i] = node
def erase(self, num: int) -> bool:
curr = self.head
ok = False
for i in range(self.level - 1, -1, -1):
curr = self.find_closest(curr, i, num)
if curr.next[i] and curr.next[i].val == num:
curr.next[i] = curr.next[i].next[i]
ok = True
while self.level > 1 and self.head.next[self.level - 1] is None:
self.level -= 1
return ok
def find_closest(self, curr: Node, level: int, target: int) -> Node:
while curr.next[level] and curr.next[level].val < target:
curr = curr.next[level]
return curr
def random_level(self) -> int:
level = 1
while level < self.max_level and random.random() < self.p:
level += 1
return level
# Your Skiplist object will be instantiated and called as such:
# obj = Skiplist()
# param_1 = obj.search(target)
# obj.add(num)
# param_3 = obj.erase(num)
Complexity
The time complexity is O(n). The space complexity is .
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.