Circular Linked List
Detailed guide and Python implementation for the 'Circular Linked List' problem.
1. Concept Overview
The 'Circular Linked List' problem is a key challenge in the Linked Lists section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Circular Linked List.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Circular Linked List carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function create_circular_linked_list(arr) that takes a list of integers arr, constructs a singly circular linked list (where the tail's next points back to the head), and returns a list of values traversing the list starting at the head and ending when the cycle is detected (each node should be visited exactly once).
- •0 <= len(arr) <= 1000
- •-10^4 <= arr[i] <= 10^4
Examples
arr = [1, 2, 3]
[1, 2, 3]
We build 1 -> 2 -> 3 -> 1. The traversal ends before visiting 1 a second time.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def create_circular_linked_list_opt(arr: list) -> list:
if not arr: return []
head = ListNode(arr[0])
curr = head
for val in arr[1:]:
curr.next = ListNode(val)
curr = curr.next
curr.next = head
res = []
c = head
visited = set()
while c and c not in visited:
res.append(c.val)
visited.add(c)
c = c.next
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def create_circular_linked_list_brute(arr: list) -> list:
if not arr: return []
head = ListNode(arr[0])
curr = head
for x in arr[1:]:
curr.next = ListNode(x)
curr = curr.next
curr.next = head
res = []
c = head
while c:
res.append(c.val)
c = c.next
if c == head:
break
return resAlgorithm Pattern Checklist
When dealing with Linked Lists data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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