Delete a Node from Circular Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the next node in the circular linked list.

DSA Python

current = head
current = current[1]

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

A.prev

B.child

C.next

D.parent

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.prev' instead of '.next' to move forward.

Using attributes that don't exist like '.child' or '.parent'.

In a circular linked list, each node points to the next node using the 'next' attribute.

2fill in blank

medium

Complete the code to check if the list has only one node.

DSA Python

if head[1] head.next:

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

A<

B!=

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' which checks for inequality instead of equality.

Using comparison operators like '>' or '<' which are not valid for node references.

If head is equal to head.next, it means the list has only one node pointing to itself.

3fill in blank

hard

Fix the error in the code to correctly delete the head node from the circular linked list.

DSA Python

head.data = head.next.data
head.next = head[1]

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

A.prev

B.next.next

C.next

D.next.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Linking head.next to head.next which causes no change.

Using '.prev' which is not relevant here.

Using '.next.prev' which is invalid in this context.

To delete the head node, copy data from the next node and link head.next to head.next.next to skip the deleted node.

4fill in blank

hard

Fill both blanks to correctly find the node before the one to delete and update links.

DSA Python

prev = head
while prev.next[1] != current:
    prev = prev.next
prev.next = current[2]

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

A.next

B.prev

C.next.next

D.next.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.prev' which is not used in singly linked circular lists.

Linking prev.next to current.prev which is invalid.

Incorrect loop condition causing infinite loops.

To find the node before current, move while prev.next != current. Then link prev.next to current.next to remove current.

5fill in blank

hard

Fill all three blanks to complete the function that deletes a node with given key from circular linked list.

DSA Python

def delete_node(head, key):
    if head is None:
        return None
    current = head
    prev = None
    while True:
        if current.data == [1]:
            if prev is not None:
                prev.next = current[2]
            else:
                head = current[3]
            return head
        prev = current
        current = current.next
        if current == head:
            break
    return head

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

Akey

B.next

Dhead.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names for key comparison.

Not updating head when deleting the head node.

Incorrectly linking nodes causing broken list.

Check if current.data equals key. If deleting non-head node, link prev.next to current.next. If deleting head, update head to head.next.

Practice

1. What happens when you delete the head node from a circular linked list with multiple nodes?

easy

A. The head moves to the next node and the list remains circular

B. The list becomes empty

C. The last node is deleted instead

D. The list breaks and becomes linear

Solution

1. Step 1: Understand the head deletion in circular linked list

   When the head node is deleted, the next node becomes the new head.

2. Step 2: Maintain circular link

   The last node's next pointer is updated to point to the new head, keeping the list circular.

3. Final Answer:

   The head moves to the next node and the list remains circular -> Option A

4. Quick Check:

   Head deletion updates head and keeps circular [OK]

Hint: Deleting head shifts head to next, list stays circular [OK]

Common Mistakes:

• Assuming list becomes empty after head deletion
• Not updating last node's next pointer
• Breaking circular link accidentally

2. Which of the following is the correct syntax to delete a node with value `key` from a circular linked list in Python?

easy

A. def deleteNode(head, key):
if head.data == key:
# delete head logic

B. def deleteNode(head, key):
while head.data != key:
head = head.next

C. def deleteNode(head, key):
if head is None:
return None

D. def deleteNode(head, key):
head = None
return head

Solution

1. Step 1: Check for empty list

   Before deleting, check if the list is empty (head is None) to avoid errors.

2. Step 2: Return None if list is empty

   If head is None, return None as there is nothing to delete.

3. Final Answer:

   def deleteNode(head, key): if head is None: return None -> Option C

4. Quick Check:

   Empty list check is essential before deletion [OK]

Hint: Always check if list is empty before deleting [OK]

Common Mistakes:

• Not checking if head is None before deletion
• Incorrectly looping without exit condition
• Setting head to None without condition

3. What will be the output list after deleting node with value 3 from this circular linked list: 1 -> 3 -> 5 -> 7 -> (back to 1)?

medium

A. 1 -> 5 -> 7 -> 3 -> (back to 1)

B. 1 -> 5 -> 7 -> (back to 1)

C. 1 -> 3 -> 5 -> 7 -> (back to 1)

D. 3 -> 5 -> 7 -> (back to 3)

Solution

1. Step 1: Locate node with value 3

   Traverse the list to find node with value 3 between 1 and 5.

2. Step 2: Remove node 3 and link 1 to 5

   Update node 1's next pointer to node 5, skipping node 3.

3. Final Answer:

   1 -> 5 -> 7 -> (back to 1) -> Option B

4. Quick Check:

   Deleted 3, linked 1 to 5 [OK]

Hint: Skip the node with key, link previous to next [OK]

Common Mistakes:

• Not updating previous node's next pointer
• Deleting wrong node or not deleting at all
• Breaking circular link after deletion

4. Given this code snippet to delete a node from a circular linked list, what is the bug?

def deleteNode(head, key):
    if head is None:
        return None
    curr = head
    while curr.next != head:
        if curr.next.data == key:
            curr.next = curr.next.next
            return head
        curr = curr.next
    return head

medium

A. Does not handle deletion if head node contains the key

B. Infinite loop due to missing break

C. Deletes all nodes instead of one

D. Returns None instead of head

Solution

1. Step 1: Analyze deletion logic

   The code checks nodes after head but never checks if head itself has the key.

2. Step 2: Identify missing head deletion handling

   If head contains the key, this code will not delete it, causing incorrect behavior.

3. Final Answer:

   Does not handle deletion if head node contains the key -> Option A

4. Quick Check:

   Head node deletion must be handled separately [OK]

Hint: Always check if head node matches key before loop [OK]

Common Mistakes:

• Ignoring head node in deletion logic
• Assuming loop covers all nodes including head
• Not updating head pointer after deletion

5. You have a circular linked list with only one node containing value 10. What should be the list after deleting node with value 10?

hard

A. List with node 10 pointing to None

B. List with node 10 pointing to itself

C. List unchanged

D. Empty list (None)

Solution

1. Step 1: Identify single node list condition

   When the list has only one node, head.next points to head itself.

2. Step 2: Delete the only node

   Deleting this node means the list becomes empty, so head should be None.

3. Final Answer:

   Empty list (None) -> Option D

4. Quick Check:

   Single node deletion empties the list [OK]

Hint: Deleting sole node results in empty list (None) [OK]

Common Mistakes:

• Leaving node pointing to itself after deletion
• Not setting head to None
• Assuming list still has one node