Detect if a Linked List is Circular in DSA C# - Time & Space Complexity

Time Complexity: Detect if a Linked List is Circular

O(n)

Understanding Time Complexity

We want to understand how long it takes to check if a linked list loops back on itself.

The question is: how does the time grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public bool IsCircular(Node head) {
    if (head == null) return false;
    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;
}
    

This code uses two pointers moving at different speeds to detect a loop in the linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves two pointers through the list.
• How many times: At most, it runs once per node until pointers meet or list ends.

How Execution Grows With Input

As the list grows, the number of steps grows roughly in proportion to the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to detect a loop grows in direct proportion to the number of nodes in the list.

Common Mistake

[X] Wrong: "The fast pointer makes this algorithm run twice as fast, so it's O(n/2)."

[OK] Correct: We drop constants in Big-O, so O(n/2) is still O(n). The main factor is the list size, not the speed difference.

Interview Connect

Understanding this time complexity helps you explain how efficient your loop detection is, a common question in interviews.

Self-Check

"What if we used only one pointer and checked visited nodes with extra memory? How would the time complexity change?"

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