Linked List Cycle

To determine if a linked list has a cycle, you can use the Floyd’s Tortoise and Hare (Cycle Detection) algorithm. It uses two pointers that move through the list at different speeds. If there is a cycle, the two pointers will eventually meet. Here’s the code:

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class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow = head
        fast = head

        while fast is not None and fast.next is not None:
            slow = slow.next
            fast = fast.next.next

            if slow == fast:
                return True

        return False

Explanation

  • Initialize two pointers, slow and fast, both starting at the head of the linked list.
  • Move slow one step at a time and fast two steps at a time.
  • If there is a cycle, slow and fast will eventually meet, and the function returns True.
  • If there is no cycle, fast will reach the end of the list, and the function returns False.

Key Takeaway

The Floyd’s Tortoise and Hare algorithm efficiently detects cycles in a linked list by using two pointers moving at different speeds. It meets the given constraints and handles the edge cases of an empty list or a list with one node without any issues.

Identifying Problem Isomorphism

“Linked List Cycle” is approximately isomorphic to the problem “Happy Number”.

Reasoning:

Both problems share a core concept, which is detecting a cycle in a sequence. In “Linked List Cycle”, the cycle is in the links between nodes in a linked list. In “Happy Number”, the cycle is detected in the sequence of numbers generated by the algorithm.

Another similar problem is “Floyd’s Cycle Detection Algorithm”.

Reasoning:

This is not a specific problem but a widely used algorithm to detect cycles in sequences or linked structures. It shares the same concept with the “Linked List Cycle” problem and is, in fact, a general method to solve it.

The order from simple to more complex is:

  1. Linked List Cycle
  2. Happy Number
  3. Floyd’s Cycle Detection Algorithm

“Linked List Cycle” is simpler as it’s more straightforward with just determining if a cycle exists in a linked list. “Happy Number” involves more complex calculations to determine if a number is happy, and the cycle detection is used as part of the overall algorithm. “Floyd’s Cycle Detection Algorithm” is the most complex as it’s a general algorithm that can be applied to various data structures and scenarios, and understanding it involves a deeper understanding of the cycle detection concept.

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class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if not head:
            return False
        slow = head
        fast = head.next
        while slow != fast:
            if not fast or not fast.next:
                return False
            slow = slow.next
            fast = fast.next.next
        return True

Problem Classification

The problem falls under the category of “Linked List Cycle Detection”. In this type of problem, the goal is to determine whether a linked list contains a cycle, which means some nodes are connected in such a way that they form a loop. The specific terminology used to describe the problem relates to the data structure “Linked List” and the concept of “Cycle Detection”. This problem is typically solved using techniques related to “Pointer Manipulation” and “Two-Pointer Techniques”, more specifically, the “Fast and Slow Pointer” or “Runner Technique”.

Language Agnostic Coding Drills

This involves determining whether a linked list has a cycle. Here’s how you can break it down into its fundamental components and solve it step by step:

  1. Understanding Linked Lists: Linked lists are a basic data structure that consists of a series of nodes, where each node contains a reference to the next node in the sequence. Understanding how to traverse and manipulate linked lists is a prerequisite to solving this problem.

  2. Understanding Pointers: Pointers or references are used to access nodes within the linked list. You should understand how to use pointers to access and manipulate linked list nodes.

  3. Understanding the ‘Fast and Slow’ Pointers Concept: This problem requires the use of two pointers moving at different speeds, commonly referred to as ‘fast’ and ‘slow’ pointers. You need to understand that the fast pointer moves two steps at a time while the slow pointer moves one step at a time.

  4. Identifying a Cycle: A cycle in a linked list is a situation where traversing the list indefinitely will result in never reaching the end because some node in the list points back to a previous node, creating a loop. The challenge is to detect the presence of such a cycle. If the fast pointer ever equals the slow pointer (they reference the same node), then a cycle exists.

  5. Handling Edge Cases: It’s important to handle cases where the linked list is empty (the head is null) or the list has only one node. The code needs to return false in these scenarios because a cycle cannot exist in these cases.

  6. Putting it All Together: Once you understand these individual components, you can combine them to solve the problem. Start both pointers at the head of the list. Move the pointers at their respective speeds. If a cycle exists, the fast pointer will eventually catch up to the slow pointer. If there’s no cycle, the fast pointer will reach the end of the list.

The key idea in this problem is the use of two pointers moving at different speeds to detect a cycle. The slow pointer advances one node at a time, while the fast pointer advances two nodes at a time. If a cycle exists, the fast pointer will eventually catch up to the slow pointer.

Targeted Drills in Python

  1. Understanding Linked Lists: Define a basic LinkedList Node structure. We will use this throughout the drills.
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class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
  1. Understanding Pointers: Let’s create two nodes and a pointer to traverse.
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node1 = ListNode(1)
node2 = ListNode(2)
node1.next = node2  # This creates a link from node1 to node2

# Let's traverse
pointer = node1
while pointer:
    print(pointer.val)
    pointer = pointer.next
  1. Understanding the ‘Fast and Slow’ Pointers Concept: Let’s use the previous nodes and add more, then use slow and fast pointers to traverse.
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node3 = ListNode(3)
node4 = ListNode(4)
node2.next = node3
node3.next = node4

slow_pointer = node1
fast_pointer = node1

while fast_pointer and fast_pointer.next:
    slow_pointer = slow_pointer.next
    fast_pointer = fast_pointer.next.next
    print(f"Slow: {slow_pointer.val}, Fast: {fast_pointer.val}")
  1. Identifying a Cycle: For this, let’s create a cycle and check if slow and fast pointers meet.
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# Creating a cycle
node4.next = node2

slow_pointer = node1
fast_pointer = node1

while fast_pointer and fast_pointer.next:
    slow_pointer = slow_pointer.next
    fast_pointer = fast_pointer.next.next
    if slow_pointer == fast_pointer:
        print("Cycle Detected")
        break

  1. Handling Edge Cases: The code already handles the case when the linked list is empty (the head is null) or when the list has only one node. No additional drills are required.

  2. Putting it All Together: Now, integrate the above steps into a function to detect a cycle in a linked list.

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class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if not head:
            return False
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False

Each of these drills build up on the previous one and when combined together, they form the complete solution.