Detect if a Linked List is Circular in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does it mean for a linked list to be circular?

A linked list is circular if its last node points back to one of the previous nodes, creating a loop instead of ending with null.

beginner

Which two pointers are commonly used to detect a cycle in a linked list?

The 'slow' pointer moves one step at a time, and the 'fast' pointer moves two steps at a time. If they meet, the list is circular.

intermediate

Why does the fast pointer moving two steps help detect a cycle?

Because if there is a cycle, the fast pointer will eventually catch up to the slow pointer inside the loop, proving the list is circular.

intermediate

What is the time complexity of detecting a cycle in a linked list using the two-pointer method?

The time complexity is O(n), where n is the number of nodes in the list, because each pointer moves through the list at most once.

beginner

Show the base case check in code to quickly return false if the linked list is empty or has only one node.

if (head == null || head.Next == null) return false; // No cycle possible

What does the fast pointer do in the cycle detection algorithm?

A Moves two steps at a time

B Moves one step at a time

C Moves three steps at a time

D Does not move

The fast pointer moves two steps at a time to catch the slow pointer if a cycle exists.

If a linked list has no cycle, what will the fast pointer eventually reach?

A Null (end of the list)

B A node pointing to itself

C The slow pointer

D The head node

If there is no cycle, the fast pointer will reach the end of the list, which is null.

What is the main idea behind Floyd’s Cycle Detection Algorithm?

A Use a hash set to store visited nodes

B Reverse the linked list

C Use two pointers moving at different speeds

D Count the number of nodes

Floyd’s algorithm uses two pointers moving at different speeds to detect cycles.

What should you check first before running the cycle detection loop?

A If the list has more than 10 nodes

B If the head or head.Next is null

C If the list is sorted

D If the list contains duplicate values

If the list is empty or has only one node, it cannot have a cycle.

What happens if the slow and fast pointers meet during traversal?

A The list has no cycle

B The list is empty

C The list is sorted

D The list is circular

Meeting of slow and fast pointers means a cycle exists in the list.

Explain how to detect if a linked list is circular using two pointers.

Think about two runners on a track at different speeds.

Describe the base cases to check before starting cycle detection in a linked list.

Consider very short or empty lists.

Practice

1. What does it mean if a linked list is circular?

easy

A. The linked list nodes are sorted in ascending order

B. The linked list has no nodes

C. The last node points back to one of the previous nodes, forming a loop

D. The linked list contains only one node

Solution

1. Step 1: Understand linked list structure

   A linked list is a chain of nodes where each node points to the next node.

2. Step 2: Define circular linked list

   If the last node points back to any previous node, it creates a loop, making the list circular.

3. Final Answer:

   The last node points back to one of the previous nodes, forming a loop -> Option C

4. Quick Check:

   Circular linked list = loop in nodes [OK]

Hint: Circular means last node links back inside list [OK]

Common Mistakes:

• Thinking circular means sorted list
• Confusing empty list with circular
• Assuming single node list is always circular

2. Which of the following C# code snippets correctly checks if a linked list is circular using two pointers?

easy

A. if (head == null) return true; else return false;

B. while (slow != null) { slow = slow.Next; if (slow == null) return true; } return false;

C. for (int i = 0; i < length; i++) { if (list[i] == list[length - 1]) return true; } return false;

D. while (fast != null && fast.Next != null) { slow = slow.Next; fast = fast.Next.Next; if (slow == fast) return true; } return false;

Solution

1. Step 1: Identify Floyd's cycle detection logic

   Two pointers move at different speeds: slow moves one step, fast moves two steps.

2. Step 2: Check for pointer meeting

   If slow and fast meet, a cycle exists; otherwise, if fast reaches null, no cycle.

3. Final Answer:

   while (fast != null && fast.Next != null) { slow = slow.Next; fast = fast.Next.Next; if (slow == fast) return true; } return false; -> Option D

4. Quick Check:

   Two-pointer cycle detection = while (fast != null && fast.Next != null) { slow = slow.Next; fast = fast.Next.Next; if (slow == fast) return true; } return false; [OK]

Hint: Use two pointers moving at different speeds [OK]

Common Mistakes:

• Checking only one pointer for null
• Using for loop on linked list nodes
• Returning true when list is empty

3. What will be the output of the following C# code snippet?

class Node { public int Data; public Node Next; public Node(int d) { Data = d; Next = null; } }
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Next = new Node(3);
head.Next.Next.Next = head.Next;
bool IsCircular(Node node) {
  Node slow = node, fast = node;
  while (fast != null && fast.Next != null) {
    slow = slow.Next;
    fast = fast.Next.Next;
    if (slow == fast) return true;
  }
  return false;
}
Console.WriteLine(IsCircular(head));

medium

A. true

B. false

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Analyze linked list structure

   Nodes 1 -> 2 -> 3, then 3 points back to node 2, creating a loop.

2. Step 2: Trace Floyd's cycle detection

   Slow and fast pointers will meet inside the loop, returning true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Loop detected = true [OK]

Hint: Loop means pointers meet inside list [OK]

Common Mistakes:

• Assuming no loop because list looks linear
• Confusing pointer equality with data equality
• Expecting false because of missing null check

4. Identify the bug in this C# method that tries to detect if a linked list is circular:

bool IsCircular(Node head) {
  Node slow = head, fast = head;
  while (fast != null) {
    slow = slow.Next;
    fast = fast.Next.Next;
    if (slow == fast) return true;
  }
  return false;
}

medium

A. Missing check for fast.Next being null before accessing fast.Next.Next

B. Slow pointer should move two steps instead of one

C. Should return false inside the loop instead of after

D. Head node should be initialized to null

Solution

1. Step 1: Review pointer movement and null checks

   fast moves two steps: fast = fast.Next.Next; must check fast.Next != null to avoid null reference.

2. Step 2: Identify missing null check

   Loop condition only checks fast != null, but fast.Next could be null causing runtime error.

3. Final Answer:

   Missing check for fast.Next being null before accessing fast.Next.Next -> Option A

4. Quick Check:

   Check fast.Next != null before fast.Next.Next [OK]

Hint: Always check fast and fast.Next for null before moving fast [OK]

Common Mistakes:

• Ignoring fast.Next null check
• Moving slow pointer two steps incorrectly
• Returning false too early inside loop

5. Given a linked list where some nodes may form multiple loops, which approach correctly detects if the list is circular and why?

public bool IsCircular(Node head) {
  Node slow = head, fast = head;
  while (fast != null && fast.Next != null) {
    slow = slow.Next;
    fast = fast.Next.Next;
    if (slow == fast) return true;
  }
  return false;
}

hard

A. It fails because it only detects the first loop and misses others

B. Floyd's cycle detection works because fast pointer will eventually meet slow pointer in any loop

C. It requires extra memory to store visited nodes to detect multiple loops

D. It only works if the list has exactly one loop

Solution

1. Step 1: Understand Floyd's cycle detection behavior

   Fast and slow pointers move at different speeds; if any loop exists, they will meet inside it.

2. Step 2: Consider multiple loops scenario

   In a singly linked list, multiple loops cannot exist independently; any loop causes pointers to meet.

3. Final Answer:

   Floyd's cycle detection works because fast pointer will eventually meet slow pointer in any loop -> Option B

4. Quick Check:

   Any loop causes pointer meeting = true [OK]

Hint: One loop detection suffices; multiple loops can't exist singly [OK]

Common Mistakes:

• Thinking multiple loops can exist in singly linked list
• Believing extra memory is needed for loop detection
• Assuming method detects all loops separately