Insert at Beginning of Circular Linked List in DSA Go - Time & Space Complexity

Time Complexity: Insert at Beginning of Circular Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed changes when we insert a new node at the start of a circular linked list.

How does the number of steps grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Insert new node at beginning of circular linked list
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
}
    

This code adds a new node at the start of a circular linked list by first finding the last node, then linking the new node before the current head.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop to find the last node (where next points back to head).
• How many times: This loop runs once for each node in the list until it reaches the last node, so it runs n times for a list of size n.

How Execution Grows With Input

As the list grows, the loop to find the last node takes longer because it checks each node once.

Input Size (n)	Approx. Operations
10	About 10 steps to find last node
100	About 100 steps to find last node
1000	About 1000 steps to find last node

Pattern observation: The number of steps grows directly with the size of the list.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert at the beginning grows linearly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Inserting at the beginning is always O(1) because we just change a few pointers."

[OK] Correct: In a circular linked list, we must find the last node to update its pointer, which takes time proportional to the list size, so it is not constant time.

Interview Connect

Understanding this helps you explain how linked list operations work in real systems, showing you can analyze and improve data structure efficiency.

Self-Check

"What if we kept a pointer to the last node? How would the time complexity of inserting at the beginning change?"

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