algoadvance

Clarifying Questions

  1. Input Clarification:
    • Are the values in the linked list nodes unique?
    • What should be returned if the linked list is empty?
  2. Cycle Details:
    • Should the solution contain advanced cycle detection techniques like Floyd’s Tortoise and Hare algorithm?
  3. Constraints:
    • Maximum length of the linked list?
    • Expected time and space complexity?

Assuming standard inputs and constraints as per typical LeetCode problems:

Strategy

We can use Floyd’s Tortoise and Hare Algorithm to detect the start of the cycle in the linked list. The main idea is as follows:

  1. Use two pointers, slow and fast. slow moves one step at a time while fast moves two steps at a time.
  2. If there is a cycle, slow and fast will eventually meet inside the cycle.
  3. Once they meet, to find the start of the cycle:
    1. Move slow to the head of the list.
    2. Move both slow and fast one step at a time; the node where they meet now is the start of the cycle.

Code

Let’s implement this in Python:

# Definition for singly-linked list node.
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, fast = head, head

    # Phase 1: Determine whether a cycle is present.
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow == fast:
            # Cycle detected
            break
    else:
        # No cycle detected
        return None

    # Phase 2: Find the start of the cycle.
    slow = head
    while slow != fast:
        slow = slow.next
        fast = fast.next
    
    return slow

Time Complexity

The time complexity of this approach is O(n), where n is the number of nodes in the linked list:

  1. Phase 1 (Cycle Detection): Both pointers traverse at most O(n) steps combined.
  2. Phase 2 (Find Cycle Start): Both pointers traverse at most O(n) steps.

Space Complexity

The space complexity is O(1) because we only use a fixed amount of additional space (pointers).

By following this strategy and implementation, we ensure an efficient and clear solution to the problem.

Try our interview co-pilot at AlgoAdvance.com