Leetcode #2674: Split a Circular Linked List
In this guide, we solve Leetcode #2674 Split a Circular Linked List 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
Given a circular linked list list of positive integers, your task is to split it into 2 circular linked lists so that the first one contains the first half of the nodes in list (exactly ceil(list.length / 2) nodes) in the same order they appeared in list, and the second one contains the rest of the nodes in list in the same order they appeared in list. Return an array answer of length 2 in which the first element is a circular linked list representing the first half and the second element is a circular linked list representing the second half.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Linked List, Two Pointers
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: nums = [1,5,7]
Output: [[1,5],[7]]
Explanation: The initial list has 3 nodes so the first half would be the first 2 elements since ceil(3 / 2) = 2 and the rest which is 1 node is in the second half.
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def splitCircularLinkedList(
self, list: Optional[ListNode]
) -> List[Optional[ListNode]]:
a = b = list
while b.next != list and b.next.next != list:
a = a.next
b = b.next.next
if b.next != list:
b = b.next
list2 = a.next
b.next = list2
a.next = list
return [list, list2]
Complexity
The time complexity is , where is the length of the linked list. 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.