Insert at End of Circular Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given data.

DSA Python

new_node = Node([1])

Drag options to blanks, or click blank then click option'

Adata

Bhead

CNone

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'head' or 'tail' instead of the data value.

Passing None instead of the data.

We create a new node by passing the data value to the Node constructor.

2fill in blank

medium

Complete the code to check if the list is empty.

DSA Python

if head is [1]:

Drag options to blanks, or click blank then click option'

Atail

Bnew_node

CNone

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if head is tail instead of None.

Using assignment '=' instead of comparison 'is'.

An empty list has head equal to None.

3fill in blank

hard

Fix the error in linking the new node to the list when the list is not empty.

DSA Python

new_node.next = [1].next
[1].next = new_node
head = new_node.next

Drag options to blanks, or click blank then click option'

Ahead

Btail

CNone

Dnew_node

Attempts:

3 left

💡 Hint

Common Mistakes

Using head instead of tail to link the new node.

Setting new_node.next to None.

We link the new node after the tail, so we use tail.next and update tail.next to new_node.

4fill in blank

hard

Fill both blanks to update the tail pointer and maintain circularity.

DSA Python

tail = [1]
tail.next = [2]

Drag options to blanks, or click blank then click option'

Anew_node

Bhead

Ctail

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating tail to new_node.

Setting tail.next to None breaking circularity.

After insertion, tail becomes the new node, and tail.next points to head to keep the list circular.

5fill in blank

hard

Fill all three blanks to complete the insert_at_end function for a circular linked list.

DSA Python

def insert_at_end(head, data):
    new_node = Node([1])
    if head is None:
        new_node.next = new_node
        head = new_node
    else:
        tail = head
        while tail.next != [2]:
            tail = tail.next
        new_node.next = tail.next
        tail.next = new_node
        tail = [3]
        tail.next = head
    return head

Drag options to blanks, or click blank then click option'

Adata

Bhead

Cnew_node

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using tail instead of head in the while loop condition.

Not updating tail to new_node after insertion.

Passing wrong value to Node constructor.

We create new_node with data, find tail by checking tail.next != head, then insert new_node after tail and update tail to new_node.

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