Detect if a Linked List is Circular in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start traversing the linked list from the head node.

DSA C#

Node current = [1];

Drag options to blanks, or click blank then click option'

Atail

Bhead

Cnull

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from tail instead of head.

Assigning null which means no node.

We start checking from the head node of the linked list.

2fill in blank

medium

Complete the condition to continue traversing while the current node is not null.

DSA C#

while (current [1] null) {
    current = current.next;
}

Drag options to blanks, or click blank then click option'

A!=

B==

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which stops immediately if current is not null.

Using comparison operators like '<' or '>' which are invalid for objects.

We continue looping while the current node exists (is not null).

3fill in blank

hard

Fix the error in the loop condition to detect if the linked list is circular using two pointers.

DSA C#

while (fast != null && fast.next [1] null) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow == fast) {
        return true;
    }
}
return false;

Drag options to blanks, or click blank then click option'

A!=

B==

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' causing runtime errors.

Not checking fast.next leading to exceptions.

The fast pointer must have a next node to jump two steps ahead safely, so we check if fast.next == null to stop.

4fill in blank

hard

Fill both blanks to complete the method that detects if a linked list is circular using Floyd's cycle detection.

DSA C#

public bool IsCircular(Node head) {
    Node slow = head;
    Node fast = head;
    while (fast != null && fast.next [1] null) {
        slow = slow.next;
        fast = fast.next.[2];
        if (slow == fast) {
            return true;
        }
    }
    return false;
}

Drag options to blanks, or click blank then click option'

A==

Bnext

C!=

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' in the loop condition.

Using fast.next.prev which is invalid in singly linked lists.

The loop continues while fast.next is not null (fast.next != null). The fast pointer moves two steps ahead using fast.next.next.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps nodes to their values only if the list is circular.

DSA C#

var nodeValues = new Dictionary<Node, int>();
Node current = head;
while (current != null) {
    if (current == [1]) break;
    nodeValues[[2]] = current.value;
    current = current.[3];
}

Drag options to blanks, or click blank then click option'

Ahead

Bcurrent

Cnext

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using tail instead of head for circular check.

Using head as dictionary key instead of current.

Using current.prev which doesn't exist in singly linked lists.

We stop if current reaches head again (circular). We use current as key and move to current.next.

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