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

Insert at Beginning of Circular Linked List

📖 Scenario: You are managing a circular linked list that stores numbers in a circle. You want to add a new number at the start of this circle so that it becomes the new first number.

🎯 Goal: Build a Go program that creates a circular linked list, sets up a new number to add, inserts it at the beginning of the list, and then prints the list to show the new order.

📋 What You'll Learn

Create a circular linked list with nodes containing integers 10, 20, 30 in that order

Create a variable called newData with the value 5

Write a function insertAtBeginning that inserts a new node with newData at the start of the circular linked list

Print the circular linked list starting from the new head node, showing all nodes once

💡 Why This Matters

🌍 Real World

Circular linked lists are used in real-time systems like music playlists or round-robin schedulers where the list loops continuously.

💼 Career

Understanding circular linked lists helps in jobs involving system programming, embedded systems, and designing efficient data structures.

1

Create Circular Linked List

Create a struct type called Node with fields data int and next *Node. Then create three nodes with data 10, 20, and 30. Link them in a circle so that the last node points back to the first node. Finally, create a variable head that points to the first node.

DSA Go

// Define Node struct and create circular linked list with 10 -> 20 -> 30 -> back to 10
// Your code here

Hint

Think of each node as a box holding a number and a pointer to the next box. Link the last box back to the first to make a circle.

2

Set New Data to Insert

Create a variable called newData and set it to the integer value 5.

DSA Go

package main

import "fmt"

type Node struct {
    data int
    next *Node
}

func main() {
    node1 := &Node{data: 10}
    node2 := &Node{data: 20}
    node3 := &Node{data: 30}

    node1.next = node2
    node2.next = node3
    node3.next = node1

    head := node1

    // Create newData variable here
    // Your code here
}

Hint

Just create a simple variable to hold the new number you want to add.

3

Insert New Node at Beginning

Write a function called insertAtBeginning that takes head *Node and data int as parameters and returns the new head *Node. This function should create a new node with the given data and insert it at the start of the circular linked list. Update the last node's next pointer to the new node. Then call this function with head and newData and assign the result back to head.

DSA Go

func insertAtBeginning(head *Node, data int) *Node {
    newNode := &Node{data: data}
    if head == nil {
        newNode.next = newNode
        return newNode
    }
    temp := head
    for temp.next != head {
        temp = temp.next
    }
    temp.next = newNode
    newNode.next = head
    return newNode
    head = insertAtBeginning(head, newData)
    // Print the circular linked list here

Hint

Find the last node by looping until its next points to head. Then link last node to new node, new node to head, and return new node as new head.

4

Print Circular Linked List

Write a loop to print the circular linked list starting from head. Print each node's data followed by ->. Stop after printing all nodes once and print null at the end.

DSA Go

Hint

Start from head and keep moving to next node, printing data. Stop when you reach head again.

Practice

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

easy

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

B. The new node is added at the end and does not affect the head.

C. The list breaks and becomes a linear linked list.

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

Solution

1. Step 1: Understand the circular linked list structure

   A circular linked list has its last node pointing back to the head, 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 A

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:

• Not updating last node's next pointer
• Breaking the circular link
• Adding node at end instead of beginning

2. Which of the following Go code snippets correctly inserts a new node at the beginning of a circular linked list?

easy

A. newNode.next = nil
head = newNode

B. head.next = newNode
newNode.next = head

C. newNode.next = head
last.next = newNode
head = newNode

D. last.next = head
head = newNode
newNode.next = last

Solution

1. Step 1: Link new node to current head

   Assign newNode.next = head to point new node to the current head.

2. Step 2: Update last node to point to new node and update head

   Set last.next = newNode to maintain circular link, then update head = newNode.

3. Final Answer:

   newNode.next = head last.next = newNode head = newNode -> Option C

4. Quick Check:

   Correct pointer updates = B [OK]

Hint: Link new node to head, update last node, then head [OK]

Common Mistakes:

• Setting newNode.next to nil breaks circular link
• Not updating last.next to new node
• Incorrect order of pointer assignments

3. What will be the printed output of the circular linked list after inserting node with value 10 at the beginning, given the list initially contains 20 -> 30 -> 40 -> (back to 20)?

func printList(head *Node) {
    if head == nil {
        fmt.Println("List is empty")
        return
    }
    temp := head
    for {
        fmt.Printf("%d -> ", temp.data)
        temp = temp.next
        if temp == head {
            break
        }
    }
    fmt.Println("null")
}

// Insert 10 at beginning
newNode := &Node{data: 10}
// Assume last points to node with 40
newNode.next = head
last.next = newNode
head = newNode

printList(head)

medium

A. List is empty

B. 20 -> 30 -> 40 -> 10 -> null

C. 10 -> 20 -> 30 -> 40 -> 10 -> null

D. 10 -> 20 -> 30 -> 40 -> null

Solution

1. Step 1: Insert new node 10 at beginning

   New node 10 points to old head 20, last node 40 points to new node 10, head updated to 10.

2. Step 2: Print list starting from new head

   Traverse from 10: 10 -> 20 -> 30 -> 40, then stop at head again.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> null -> Option D

4. Quick Check:

   New head 10 followed by old nodes = A [OK]

Hint: New node prints first, then old nodes in order [OK]

Common Mistakes:

• Printing extra node causing repeated values
• Not updating last.next causing infinite loop
• Confusing order of nodes after insertion

4. Given the following Go code to insert at beginning of a circular linked list, what is the error?

func insertAtBeginning(head *Node, data int) *Node {
    newNode := &Node{data: data}
    if head == nil {
        newNode.next = newNode
        return newNode
    }
    temp := head
    for temp.next != nil {
        temp = temp.next
    }
    newNode.next = head
    temp.next = newNode
    head = newNode
    return head
}

medium

A. The function should not return head.

B. The loop condition should check for temp.next != head, not nil.

C. head should not be updated inside the function.

D. newNode.next should be nil when list is empty.

Solution

1. Step 1: Analyze loop condition for circular list

   In circular linked list, last node's next points to head, so loop should check temp.next != head, not nil.

2. Step 2: Identify impact of wrong condition

   Checking temp.next != nil causes infinite loop or crash because next is never nil in circular list.

3. Final Answer:

   The loop condition should check for temp.next != head, not nil. -> Option B

4. Quick Check:

   Loop must stop at head, not nil [OK]

Hint: In circular list, last.next points to head, not nil [OK]

Common Mistakes:

• Using nil instead of head in loop condition
• Not updating head after insertion
• Returning wrong pointer from function

5. You have a circular linked list with nodes 5 -> 15 -> 25 -> (back to 5). You want to insert a new node with value 1 at the beginning. After insertion, you want to print the list starting from the new head. Which sequence of pointer updates correctly achieves this without breaking the circular structure?

hard

A. Set newNode.next = head; find last node and set last.next = newNode; update head = newNode

B. Set newNode.next = nil; update head = newNode; last.next = head

C. Update head = newNode; set newNode.next = head; last.next = newNode

D. Set last.next = newNode; newNode.next = head; head = newNode.next

Solution

1. Step 1: Link new node to current head

   Assign newNode.next = head so new node points to old head.

2. Step 2: Update last node to point to new node and update head

   Find last node (with value 25) and set last.next = newNode to maintain circular link, then update head = newNode.

3. Final Answer:

   Set newNode.next = head; find last node and set last.next = newNode; update head = newNode -> Option A

4. Quick Check:

   New head points to old head; last points to new head [OK]

Hint: Link new node to head, last to new node, then update head [OK]

Common Mistakes:

• Setting newNode.next to nil breaks circle
• Updating head before linking nodes
• Not updating last.next to new node