Insert at End of Circular Linked List in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a circular linked list?

A circular linked list is a linked list where the last node points back to the first node, forming a circle. There is no null at the end.

beginner

Why do we need a special method to insert at the end of a circular linked list?

Because the last node points to the first node, we must update the last node's next pointer to the new node and the new node's next pointer to the first node to keep the circle intact.

intermediate

In Go, what is the key step to insert a new node at the end of a circular linked list?

Find the last node (which points to the head), set its next pointer to the new node, and set the new node's next pointer to the head node.

beginner

What happens if the circular linked list is empty when inserting at the end?

The new node points to itself, making it the only node in the list, and the head points to this new node.

intermediate

Show the Go code snippet to insert a node at the end of a circular linked list.

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

In a circular linked list, what does the last node point to?

A The first node

B Null

C Itself

D The second node

In a circular linked list, the last node points back to the first node to form a circle.

When inserting at the end of a circular linked list, what must the new node's next pointer point to?

A The head node

B The last node

C Itself

D Null

The new node's next pointer must point to the head node to maintain the circular structure.

If the circular linked list is empty, how should the new node's next pointer be set when inserting?

A Point to a new node

B Null

C Point to head

D Point to itself

When the list is empty, the new node points to itself to form a single-node circular list.

What is the time complexity to insert at the end of a circular linked list if you only have the head pointer?

A O(1)

B O(n)

C O(log n)

D O(n^2)

You must traverse the list to find the last node, which takes O(n) time.

In Go, which keyword is used to create a new node dynamically?

A make

B new

C &

D var

The & operator is used to get the address of a struct literal, creating a pointer to a new node.

Explain step-by-step how to insert a new node at the end of a circular linked list.

Think about how to keep the circle intact after insertion.

Write the Go function signature and describe the logic to insert a node at the end of a circular linked list.

Focus on pointer updates and edge cases.

Practice

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

easy

A. The new node points to itself, becoming the only node in the list

B. The new node points to null, ending the list

C. The list remains empty because insertion is not possible

D. The new node points to the head, but the list has multiple nodes

Solution

1. Step 1: Understand empty circular linked list

   When the list is empty, there is no node, so the new node must point to itself to maintain circularity.

2. Step 2: Insert new node and update pointers

   The new node's next pointer is set to itself, making it the only node and the head of the list.

3. Final Answer:

   The new node points to itself, becoming the only node in the list -> Option A

4. Quick Check:

   Empty list insertion = single node pointing to itself [OK]

Hint: New node points to itself if list is empty [OK]

Common Mistakes:

• Setting new node's next to null instead of itself
• Not updating head pointer
• Assuming list remains empty after insertion

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

easy

A. newNode.next = nil
last.next = newNode
last = newNode

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

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

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

Solution

1. Step 1: Set new node's next to head

   In circular linked list, new node's next must point to head to maintain circularity.

2. Step 2: Update last node's next and last pointer

   Last node's next points to new node, and last pointer updates to new node.

3. Final Answer:

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

4. Quick Check:

   Correct pointer updates maintain circular list [OK]

Hint: New node next = head; last next = new node; update last [OK]

Common Mistakes:

• Setting newNode.next to nil breaks circularity
• Not updating last pointer to new node
• Incorrect pointer assignments causing runtime errors

3. Given the following Go code for inserting at the end of a circular linked list, what will be the printed list after inserting 4?

type Node struct {
  data int
  next *Node
}

func insertEnd(last *Node, data int) *Node {
  newNode := &Node{data: data}
  if last == nil {
    newNode.next = newNode
    return newNode
  }
  newNode.next = last.next
  last.next = newNode
  return newNode
}

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

func main() {
  var last *Node
  last = insertEnd(last, 1)
  last = insertEnd(last, 2)
  last = insertEnd(last, 3)
  last = insertEnd(last, 4)
  printList(last)
}

medium

A. 1 -> 2 -> 3 -> null

B. 4 -> 1 -> 2 -> 3 -> null

C. 2 -> 3 -> 4 -> 1 -> null

D. 1 -> 2 -> 3 -> 4 -> null

Solution

1. Step 1: Trace insertions in order

   Insert 1: list is empty, new node points to itself. Insert 2: new node added after last, last updated. Insert 3 and 4 similarly.

2. Step 2: Print list starting from last.next

   Printing starts at first node (last.next), so order is 1 -> 2 -> 3 -> 4 -> null.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> null -> Option D

4. Quick Check:

   Insertion at end maintains order [OK]

Hint: Print starts at last.next, so order is insertion order [OK]

Common Mistakes:

• Assuming last node is printed first
• Not updating last pointer after insertion
• Confusing circular list print start point

4. The following Go function is intended to insert a new node at the end of a circular linked list. What is the bug?

func insertEnd(last *Node, data int) *Node {
  newNode := &Node{data: data}
  if last == nil {
    newNode.next = nil
    return newNode
  }
  newNode.next = last.next
  last.next = newNode
  return newNode
}

medium

A. newNode.next should point to itself when list is empty, not nil

B. last.next should be set to nil instead of newNode

C. Function should return last, not newNode

D. Data field of newNode is not initialized

Solution

1. Step 1: Check empty list insertion

   When list is empty, newNode.next must point to itself to maintain circularity, not nil.

2. Step 2: Identify incorrect pointer assignment

   Setting newNode.next to nil breaks circular linked list property causing runtime errors.

3. Final Answer:

   newNode.next should point to itself when list is empty, not nil -> Option A

4. Quick Check:

   Empty list node next = itself [OK]

Hint: Empty list node next must point to itself, never nil [OK]

Common Mistakes:

• Setting next to nil breaks circularity
• Returning newNode instead of last can confuse list end
• Not initializing data field properly

5. You have a circular linked list with nodes 10 -> 20 -> 30 -> (back to 10). You want to insert 40 at the end. After insertion, which node should the last pointer refer to, and what will be the list order when printed starting from last.next? Choose the correct pair.

hard

A. Last points to node with data 30; list order: 40 -> 10 -> 20 -> 30 -> null

B. Last points to node with data 40; list order: 10 -> 20 -> 30 -> 40 -> null

C. Last points to node with data 40; list order: 40 -> 10 -> 20 -> 30 -> null

D. Last points to node with data 30; list order: 10 -> 20 -> 30 -> 40 -> null

Solution

1. Step 1: Insert new node at end and update last pointer

   New node 40 is added after last (30), so last pointer updates to new node 40.

2. Step 2: Print list starting from last.next

   last.next points to head (10), so printing order is 10 -> 20 -> 30 -> 40 -> null.

3. Final Answer:

   Last points to node with data 40; list order: 10 -> 20 -> 30 -> 40 -> null -> Option B

4. Quick Check:

   Last updated to new end node; print starts at last.next [OK]

Hint: Last always points to newest end node; print starts at last.next [OK]

Common Mistakes:

• Not updating last pointer after insertion
• Printing starting from last instead of last.next
• Confusing list order after insertion