Insert at End of Circular Linked List in DSA Python - Time & Space Complexity

Time Complexity: Insert at End of Circular Linked List

O(n)

Understanding Time Complexity

We want to understand how the time to insert a new node at the end of a circular linked list changes as the list grows.

How does the number of steps needed grow when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = new_node
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head
    

This code inserts a new node at the end of a circular linked list by traversing the list to find the last node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through each node to find the last node.
• How many times: It runs once for each node in the list until it reaches the last node.

How Execution Grows With Input

As the list gets longer, the loop runs more times because it checks each node to find the end.

Input Size (n)	Approx. Operations
10	About 10 steps to find the end
100	About 100 steps to find the end
1000	About 1000 steps to find the end

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

Final Time Complexity

Time Complexity: O(n)

This means the time to insert at the end grows linearly with the size of the list.

Common Mistake

[X] Wrong: "Inserting at the end is always constant time because we just add a node."

[OK] Correct: We must first find the last node by walking through the list, which takes time proportional to the list size.

Interview Connect

Understanding this helps you explain how linked list operations scale and shows you can analyze real data structure behaviors clearly.

Self-Check

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

Practice

1. What happens when you insert a new node at the end of a circular linked list that currently has nodes?

easy

A. The new node is ignored because the list is circular.

B. The new node becomes the head node and points to itself only.

C. The new node is inserted at the beginning and breaks the circular link.

D. The new node becomes the last node and points back to the head node.

Solution

1. Step 1: Understand the circular linked list structure

   In a circular linked list, the last node points back to the head node, forming a circle.

2. Step 2: Insert new node at the end

   The new node is linked after the current last node, and its next pointer is set to the head node, maintaining the circle.

3. Final Answer:

   The new node becomes the last node and points back to the head node. -> Option D

4. Quick Check:

   Insert at end keeps circle intact = D [OK]

Hint: New end node always points to head to keep circle [OK]

Common Mistakes:

• Forgetting to link new node back to head
• Breaking the circular link by pointing to None
• Inserting at beginning instead of end

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

easy

A. new_node = Node(data); new_node.next = None; last.next = new_node

B. new_node = Node(data); new_node.next = head; last.next = new_node

C. new_node = Node(data); head = new_node; new_node.next = head

D. new_node = Node(data); last.next = None; new_node.next = last

Solution

1. Step 1: Create new node and link to head

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

2. Step 2: Link last node to new node

   The current last node's next pointer is updated to point to the new node, completing the insertion.

3. Final Answer:

   new_node = Node(data); new_node.next = head; last.next = new_node -> Option B

4. Quick Check:

   New node points to head, last points to new node = A [OK]

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

Common Mistakes:

• Setting new node next to None breaks circularity
• Not updating last node's next pointer
• Assigning head to new node incorrectly

3. What will be the printed output of the circular linked list after inserting 4 at the end, given the initial list is 1 -> 2 -> 3 -> (back to 1)?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = new_node
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head

head = Node(1)
second = Node(2)
third = Node(3)
head.next = second
second.next = third
third.next = head

head = insert_end(head, 4)

# Print circular list
current = head
output = []
while True:
    output.append(current.data)
    current = current.next
    if current == head:
        break
print(output)

medium

A. [1, 2, 3]

B. [4, 1, 2, 3]

C. [1, 2, 3, 4]

D. [1, 2, 3, 4, 1]

Solution

1. Step 1: Insert 4 at the end of the circular list

   The function finds the last node (3), links it to new node (4), and new node points back to head (1).

2. Step 2: Print nodes starting from head

   Traverse from head: 1 -> 2 -> 3 -> 4, then back to 1 stops loop. Output list is [1, 2, 3, 4].

3. Final Answer:

   [1, 2, 3, 4] -> Option C

4. Quick Check:

   List after insert = [1, 2, 3, 4] [OK]

Hint: Traverse until back to head to print full circular list [OK]

Common Mistakes:

• Printing extra repeated head node
• Not updating last node's next pointer
• Stopping traversal too early

4. Identify the error in this code snippet for inserting at the end of a circular linked list:

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = None
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head

medium

A. new_node.next should point to new_node when list is empty

B. temp.next should be set to None instead of new_node

C. The while loop condition should be temp.next == head

D. The function should return new_node instead of head

Solution

1. Step 1: Check empty list case

   When list is empty, new_node.next must point to itself to form a circular list, not None.

2. Step 2: Verify rest of code

   Other parts correctly insert new_node at end and maintain circularity.

3. Final Answer:

   new_node.next should point to new_node when list is empty -> Option A

4. Quick Check:

   Empty list node points to itself = B [OK]

Hint: Empty circular list node points to itself, not None [OK]

Common Mistakes:

• Setting new_node.next to None breaks circularity
• Wrong while loop condition
• Returning wrong node after insertion

5. Given a circular linked list with nodes 10 -> 20 -> 30 -> (back to 10), write the output after inserting 5 at the end and then inserting 40 at the end. What is the final list printed?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = new_node
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head

head = Node(10)
second = Node(20)
third = Node(30)
head.next = second
second.next = third
third.next = head

head = insert_end(head, 5)
head = insert_end(head, 40)

current = head
output = []
while True:
    output.append(current.data)
    current = current.next
    if current == head:
        break
print(output)

hard

A. [10, 20, 30, 5, 40]

B. [5, 10, 20, 30, 40]

C. [10, 20, 30, 40, 5]

D. [40, 10, 20, 30, 5]

Solution

1. Step 1: Insert 5 at the end

   After first insertion, list becomes 10 -> 20 -> 30 -> 5 -> (back to 10).

2. Step 2: Insert 40 at the end

   After second insertion, list becomes 10 -> 20 -> 30 -> 5 -> 40 -> (back to 10).

3. Step 3: Print the list

   Traverse from head: 10, 20, 30, 5, 40, then back to 10 stops loop. Output is [10, 20, 30, 5, 40].

4. Final Answer:

   [10, 20, 30, 5, 40] -> Option A

5. Quick Check:

   Two inserts at end add nodes in order = A [OK]

Hint: Insertions add nodes at end, order preserved [OK]

Common Mistakes:

• Inserting new nodes at beginning by mistake
• Not updating last node's next pointer correctly
• Printing extra repeated nodes