Introduction

In this blog post, we will explore how to solve the LeetCode problem "142" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null. There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter. Do not modify the linked list. Example 1: Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 Explanation: There is a cycle in the linked list, where tail connects to the second node. Example 2: Input: head = [1,2], pos = 0 Output: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node. Example 3: Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list. Constraints: The number of the nodes in the list is in the range [0, 104]. -105 <= Node.val <= 105 pos is -1 or a valid index in the linked-list. Follow up: Can you solve it using O(1) (i.e. constant) memory?

Explanation

Here's the solution to detect the start of a cycle in a linked list, adhering to the given constraints and complexity requirements.

• Floyd's Cycle Detection Algorithm: Use two pointers, 'slow' and 'fast'. The 'slow' pointer moves one node at a time, while the 'fast' pointer moves two nodes at a time.
• Detect Cycle: If a cycle exists, the 'fast' pointer will eventually meet the 'slow' pointer.

• Find Cycle Start: Once the pointers meet, move one pointer to the head of the list and advance both pointers one node at a time until they meet again. The meeting point is the start of the cycle.

• Runtime Complexity: O(N), where N is the number of nodes in the list.

• Storage Complexity: O(1)

Code

    # Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

def detectCycle(head: ListNode) -> ListNode:
    if not head or not head.next:
        return None

    slow = head
    fast = head

    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next

        if slow == fast:
            # Cycle detected
            slow = head
            while slow != fast:
                slow = slow.next
                fast = fast.next
            return slow

    return None