Insert at Beginning of Circular Linked List in DSA C# - Build from Scratch

Insert at Beginning of Circular Linked List

📖 Scenario: You are managing a circular linked list that represents a round-robin schedule of tasks. You need to add a new task at the beginning of the schedule so it runs first.

🎯 Goal: Build a circular linked list and write code to insert a new node at the beginning of the list.

📋 What You'll Learn

Create a Node class with int data and Node next fields

Create a circular linked list with three nodes containing data 10, 20, and 30

Create a method InsertAtBeginning that inserts a new node at the start

Print the circular linked list starting from the head node after insertion

💡 Why This Matters

🌍 Real World

Circular linked lists are used in scheduling systems, multiplayer games, and resource sharing where the process repeats in a cycle.

💼 Career

Understanding circular linked lists helps in roles involving system design, embedded systems, and real-time applications where cyclic data structures are common.

1

Create the Circular Linked List with Three Nodes

Create a Node class with int data and Node next fields. Then create three nodes with data 10, 20, and 30. Link them to form a circular linked list where the last node points back to the first node. Store the first node in a variable called head.

DSA C#

// Define Node class and create circular linked list with nodes 10, 20, 30
// Your code here

Hint

Remember to link the last node's next to the first node to make the list circular.

2

Create a Method to Insert at Beginning

Add a static method called InsertAtBeginning that takes two parameters: Node head and int data. This method will insert a new node with the given data at the beginning of the circular linked list and return the new head node.

DSA C#

public class Node
{
    public int data;
    public Node next;

    public Node(int data)
    {
        this.data = data;
        this.next = null;
    }
}

class Program
{
    static Node InsertAtBeginning(Node head, int data)
    {
        // Your code here
    }

    static void Main()
    {
        Node node1 = new Node(10);
        Node node2 = new Node(20);
        Node node3 = new Node(30);

        node1.next = node2;
        node2.next = node3;
        node3.next = node1; // circular link

        Node head = node1;
    }
}

Hint

Find the last node to update its next to the new node, then set the new node's next to the old head.

3

Insert a New Node with Data 5 at the Beginning

In the Main method, call InsertAtBeginning with head and 5 as arguments. Update head to the returned node.

DSA C#

public class Node
{
    public int data;
    public Node next;

    public Node(int data)
    {
        this.data = data;
        this.next = null;
    }
}

class Program
{
    static Node InsertAtBeginning(Node head, int data)
    {
        Node newNode = new Node(data);
        if (head == null)
        {
            newNode.next = newNode;
            return newNode;
        }

        Node temp = head;
        while (temp.next != head)
        {
            temp = temp.next;
        }

        temp.next = newNode;
        newNode.next = head;
        return newNode;
    }

    static void Main()
    {
        Node node1 = new Node(10);
        Node node2 = new Node(20);
        Node node3 = new Node(30);

        node1.next = node2;
        node2.next = node3;
        node3.next = node1; // circular link

        Node head = node1;

        // Your code here
    }
}

Hint

Assign the result of InsertAtBeginning(head, 5) back to head.

4

Print the Circular Linked List Starting from Head

Add a PrintList method that takes Node head and prints the data of each node in the circular linked list separated by -> . Then call PrintList(head) in Main after insertion.

DSA C#

public class Node
{
    public int data;
    public Node next;

    public Node(int data)
    {
        this.data = data;
        this.next = null;
    }
}

class Program
{
    static Node InsertAtBeginning(Node head, int data)
    {
        Node newNode = new Node(data);
        if (head == null)
        {
            newNode.next = newNode;
            return newNode;
        }

        Node temp = head;
        while (temp.next != head)
        {
            temp = temp.next;
        }

        temp.next = newNode;
        newNode.next = head;
        return newNode;
    }

    static void PrintList(Node head)
    {
        // Your code here
    }

    static void Main()
    {
        Node node1 = new Node(10);
        Node node2 = new Node(20);
        Node node3 = new Node(30);

        node1.next = node2;
        node2.next = node3;
        node3.next = node1; // circular link

        Node head = node1;

        head = InsertAtBeginning(head, 5);

        // Your code here
    }
}

Hint

Use a do-while loop to print nodes until you reach the head again.

Practice

1. What happens when you insert a new node at the beginning of a circular linked list?

easy

A. The list becomes linear instead of circular.

B. The new node is added at the end, breaking the circular link.

C. The new node replaces the last node in the list.

D. The new node becomes the head and points to the old head, maintaining the circular link.

Solution

1. Step 1: Understand the circular linked list structure

   A circular linked list has its last node pointing back to the head node, forming a circle.

2. Step 2: Insert new node at beginning

   The new node is linked to the current head, and the last node's next pointer is updated to this new node, making it the new head.

3. Final Answer:

   The new node becomes the head and points to the old head, maintaining the circular link. -> Option D

4. Quick Check:

   New head points to old head, circle intact = A [OK]

Hint: New head points to old head, last node points to new head [OK]

Common Mistakes:

• Forgetting to update last node's next pointer
• Breaking the circular link by not linking last node
• Replacing the last node instead of adding new head

2. Which of the following is the correct syntax to create a new node and insert it at the beginning of a circular linked list in C#?

easy

A. Node newNode = new Node(data); newNode.next = head; head = newNode;

B. Node newNode = new Node(data); newNode.next = null; head = newNode;

C. Node newNode = new Node(data); head.next = newNode; newNode.next = head;

D. Node newNode = new Node(data); newNode.next = head.next; head = newNode;

Solution

1. Step 1: Create new node with data

   Use 'new Node(data)' to create a new node with the given data.

2. Step 2: Link new node to current head and update head

   Set newNode.next to current head, then update head to newNode to insert at beginning.

3. Final Answer:

   Node newNode = new Node(data); newNode.next = head; head = newNode; -> Option A

4. Quick Check:

   New node points to old head, head updated = C [OK]

Hint: New node next points to head, then head updated [OK]

Common Mistakes:

• Setting newNode.next to null breaks circular link
• Not updating head to newNode
• Incorrectly linking nodes causing broken list

3. Given the following circular linked list with nodes 10 -> 20 -> 30 -> back to 10, what will be the list after inserting 5 at the beginning?

Node newNode = new Node(5);
newNode.next = head;
Node temp = head;
while(temp.next != head) {
  temp = temp.next;
}
temp.next = newNode;
head = newNode;

medium

A. 5 -> 10 -> 20 -> 30 -> back to 5

B. 10 -> 20 -> 30 -> 5 -> back to 10

C. 5 -> 10 -> 20 -> 30 -> back to 10

D. 10 -> 20 -> 30 -> back to 5

Solution

1. Step 1: Link new node to current head

   newNode.next points to head (10), so newNode links to 10.

2. Step 2: Find last node and update its next pointer

   Traverse to last node (30), update its next to newNode (5), maintaining circular link.

3. Step 3: Update head to newNode

   Head now points to newNode (5), making it the first node.

4. Final Answer:

   5 -> 10 -> 20 -> 30 -> back to 5 -> Option A

5. Quick Check:

   New head 5, last node points to 5 = D [OK]

Hint: Last node must point to new head after insertion [OK]

Common Mistakes:

• Not updating last node's next pointer to new head
• Head not updated to new node
• Breaking circular link by wrong pointer assignment

4. Identify the error in the following code snippet that inserts a node at the beginning of a circular linked list:

Node newNode = new Node(data);
newNode.next = head;
head = newNode;
Node temp = head;
while(temp.next != head) {
  temp = temp.next;
}
temp.next = newNode;

medium

A. The newNode.next should be set to null.

B. The last node's next pointer is not updated correctly.

C. The head is updated before fixing the last node's next pointer, causing an infinite loop.

D. The while loop condition should be temp.next != null.

Solution

1. Step 1: Analyze head update timing

   Head is updated to newNode before finding last node, so temp starts at newNode.

2. Step 2: Check while loop condition

   temp.next (old head) != head (newNode) is true, so loop enters but traverses the old cycle indefinitely since old last.next points to old head != new head, causing infinite loop.

3. Step 3: Consequence of error

   Last node's next pointer is not updated correctly, breaking circular link or causing infinite loop.

4. Final Answer:

   The head is updated before fixing the last node's next pointer, causing an infinite loop. -> Option C

5. Quick Check:

   Update last node after head change causes loop error = A [OK]

Hint: Update last node's next before changing head [OK]

Common Mistakes:

• Changing head before fixing last node's pointer
• Incorrect loop condition for circular list
• Setting newNode.next to null breaks circle

5. You have a circular linked list with only one node (data = 10). You want to insert a new node with data = 5 at the beginning. Which of the following code snippets correctly updates the list to maintain circularity?

hard

A. Node newNode = new Node(5);
newNode.next = null;
head.next = newNode;
head = newNode;

B. Node newNode = new Node(5);
newNode.next = head;
head.next = newNode;
head = newNode;

C. Node newNode = new Node(5);
newNode.next = head.next;
head = newNode;

D. Node newNode = new Node(5);
newNode.next = head;
head = newNode;

Solution

1. Step 1: Link new node to current head

   newNode.next = head links new node to existing node (10).

2. Step 2: Update old node's next to new node

   head.next = newNode ensures the old node points back to new node, maintaining circularity.

3. Step 3: Update head to new node

   head = newNode makes new node the first node in the list.

4. Final Answer:

   Node newNode = new Node(5); newNode.next = head; head.next = newNode; head = newNode; -> Option B

5. Quick Check:

   Single node circle updated with new head and links = B [OK]

Hint: For single node, update old node's next to new node [OK]

Common Mistakes:

• Setting newNode.next to null breaks circle
• Not updating old node's next pointer
• Not updating head to new node