Insert at Beginning of Circular Linked List in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of inserting at beginning in circular linked list

What is the printed state of the circular linked list after inserting 10 at the beginning?

DSA Go

package main
import "fmt"
type Node struct {
    data int
    next *Node
}
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
}
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")
}
func main() {
    var head *Node
    head = insertAtBeginning(head, 5)
    head = insertAtBeginning(head, 10)
    printList(head)
}

A 5 -> 10 -> null

B 10 -> 5 -> null

C 10 -> null

D 5 -> null

💡 Hint

Remember that inserting at beginning means new node points to old head and last node points to new node.

Inserting 10 at beginning makes 10 the new head. The last node (which had 5) points to 10, maintaining circularity. So the list prints 10 -> 5 -> null.

🧠 Conceptual

intermediate

Understanding circular linked list insertion

Why do we need to update the last node's next pointer when inserting at the beginning of a circular linked list?

A To reverse the list order

B To delete the old head node

C To maintain the circular link so the last node points to the new head

D To break the circular link and make it linear

💡 Hint

Think about what makes the list circular.

In a circular linked list, the last node points back to the head. When inserting a new node at the beginning, the last node must point to this new node to keep the circle intact.

🔧 Debug

advanced

Identify the error in circular linked list insertion code

What error will occur if we forget to update the last node's next pointer when inserting at the beginning?

DSA Go

func insertAtBeginning(head *Node, data int) *Node {
    newNode := &Node{data: data}
    if head == nil {
        newNode.next = newNode
        return newNode
    }
    // Missing update of last node's next pointer
    newNode.next = head
    return newNode
}

A The list will become linear and lose circularity

B The program will crash with nil pointer dereference

C The new node will not be inserted

D The list will be reversed

💡 Hint

What happens to the last node's next pointer?

Without updating the last node's next pointer, it still points to the old head, breaking the circular link. The list becomes linear starting from the new node.

❓ Predict Output

advanced

Output after multiple insertions at beginning

What is the printed state of the circular linked list after inserting 3, then 7, then 1 at the beginning?

DSA Go

package main
import "fmt"
type Node struct {
    data int
    next *Node
}
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
}
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")
}
func main() {
    var head *Node
    head = insertAtBeginning(head, 3)
    head = insertAtBeginning(head, 7)
    head = insertAtBeginning(head, 1)
    printList(head)
}

A 1 -> 3 -> 7 -> null

B 3 -> 7 -> 1 -> null

C 7 -> 3 -> 1 -> null

D 1 -> 7 -> 3 -> null

💡 Hint

Each new node becomes the new head.

Inserting 3 first makes list: 3 -> null. Then inserting 7 at beginning: 7 -> 3 -> null. Then inserting 1 at beginning: 1 -> 7 -> 3 -> null.

🧠 Conceptual

expert

Why is circular linked list insertion at beginning O(n)?

Why does inserting a node at the beginning of a circular linked list take O(n) time?

A Because we must traverse the list to find the last node to update its next pointer

B Because inserting at beginning requires shifting all nodes

C Because circular linked lists have no head pointer

D Because we must reverse the list before insertion

💡 Hint

Think about what pointer needs updating to maintain circularity.

To maintain circularity, the last node's next pointer must point to the new head. Finding the last node requires traversing the list, which takes O(n) time.

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