Delete from End of Doubly Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the doubly linked list is empty before deletion.

DSA Python

if self.head is [1]:
    return None

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

ANone

BFalse

CTrue

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using True or False instead of None.

Checking if head equals 0.

We check if head is None to know if the list is empty.

2fill in blank

medium

Complete the code to move to the last node in the doubly linked list.

DSA Python

current = self.head
while current.[1] is not None:
    current = current.next

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

Atail

Bhead

Cprev

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next to move forward.

Using head or tail which are not node attributes.

We use current.next to move forward through the list until the last node.

3fill in blank

hard

Fix the error in updating the previous node's next pointer when deleting the last node.

DSA Python

if current.prev is not None:
    current.prev.[1] = None

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

Ahead

Bnext

Cprev

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev instead of next.

Trying to set head or tail which are not node attributes.

We set the previous node's next pointer to None to remove the last node.

4fill in blank

hard

Fill both blanks to handle the case when the list has only one node during deletion.

DSA Python

if current.prev is [1] and current.next is [2]:
    self.head = None

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

ANone

BTrue

CFalse

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using True or False instead of None.

Using 0 which is not a pointer.

If both prev and next are None, the list has one node, so head becomes None after deletion.

5fill in blank

hard

Fill all three blanks to complete the delete from end method in the doubly linked list.

DSA Python

def delete_from_end(self):
    if self.head is [1]:
        return None
    current = self.head
    while current.[2] is not None:
        current = current.next
    if current.prev is not [3]:
        current.prev.next = None
    else:
        self.head = None

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

ANone

BTrue

Cnext

DFalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using True or False instead of None.

Mixing up prev and next pointers.

We check if head is None, move using next, and check if prev is None to update pointers correctly.

Practice

1. What happens when you delete a node from the end of a doubly linked list that has multiple nodes?

easy

A. The last node is removed and the second last node becomes the new tail.

B. The first node is removed and the second node becomes the new head.

C. All nodes are deleted and the list becomes empty.

D. The middle node is removed and the list remains unchanged at the ends.

Solution

1. Step 1: Identify the last node and update pointers

   In a doubly linked list, the last node is the tail node which has its next pointer as null. Deleting the last node means updating the tail pointer to the previous node and setting its next pointer to null. The second last node becomes the new tail.

2. Final Answer:

   The last node is removed and the second last node becomes the new tail. -> Option A

3. Quick Check:

   Delete end node = tail moves back [OK]

Hint: Last node removed, tail moves to previous node [OK]

Common Mistakes:

• Removing the head instead of the tail
• Not updating the tail pointer
• Leaving the next pointer of new tail non-null

2. Which of the following is the correct syntax to delete the last node in a doubly linked list in Python?

easy

A. head = head.prev; head.next = None

B. tail = tail.next; tail.prev = None

C. head = head.next; head.prev = None

D. tail = tail.prev; tail.next = None

Solution

1. Step 1: Move tail pointer and update next

   To delete the last node, move tail to tail.prev (second last node), then set tail.next to None to mark the end of the list.

2. Final Answer:

   tail = tail.prev; tail.next = None -> Option D

3. Quick Check:

   Correct tail update syntax = tail = tail.prev; tail.next = None [OK]

Hint: Tail moves back, next pointer set to None [OK]

Common Mistakes:

• Moving tail forward instead of backward
• Setting prev pointer instead of next
• Changing head pointer instead of tail

3. What will be the output after deleting the last node from the doubly linked list shown below?

Initial list: 10 <-> 20 <-> 30 <-> 40 <-> None
Operation: Delete from end

What is the resulting list?

medium

A. 10 <-> 20 <-> 30 <-> 40 <-> None

B. 10 <-> 20 <-> 30 <-> None

C. 20 <-> 30 <-> 40 <-> None

D. 10 <-> 20 <-> None

Solution

1. Step 1: Identify last node and update

   The last node is 40, which is removed by setting node 30's next pointer to None, making it the new tail. Result: 10 <-> 20 <-> 30 <-> None.

2. Final Answer:

   10 <-> 20 <-> 30 <-> None -> Option B

3. Quick Check:

   Last node 40 removed, tail is 30 [OK]

Hint: Remove last node, update previous node's next to None [OK]

Common Mistakes:

• Not updating the next pointer of new tail
• Removing the wrong node (head or middle)
• Leaving dangling pointers causing errors

4. Consider the following code snippet to delete the last node from a doubly linked list:

if tail is None:
    print('List is empty')
else:
    tail = tail.prev
    tail.next = None

What error can occur when this code runs on a list with only one node?

medium

A. AttributeError because tail.prev is None and tail.next is accessed

B. No error, code works fine for single node list

C. SyntaxError due to missing colon

D. IndexError because tail is out of range

Solution

1. Step 1: Trace execution for single node

   For one node, tail.prev is None, so tail = None, then tail.next = None accesses None.next causing AttributeError.

2. Final Answer:

   AttributeError because tail.prev is None and tail.next is accessed -> Option A

3. Quick Check:

   Single node delete needs extra check [OK]

Hint: Check if tail.prev is None before setting tail.next [OK]

Common Mistakes:

• Not handling single node list case
• Assuming tail.prev always exists
• Ignoring NoneType attribute errors

5. You have a doubly linked list with nodes: 5 <-> 15 <-> 25 <-> 35 <-> 45 <-> None. You delete the last node twice in a row. What will be the final list?

hard

A. 5 <-> 15 <-> 25 <-> 35 <-> 45 <-> None

B. 5 <-> 15 <-> 25 <-> 35 <-> None

C. 5 <-> 15 <-> 25 <-> None

D. 15 <-> 25 <-> 35 <-> None

Solution

1. Step 1: Perform two deletions

   Initial: 5 <-> 15 <-> 25 <-> 35 <-> 45. First delete removes 45 (tail=35). Second delete removes 35 (tail=25). Final: 5 <-> 15 <-> 25 <-> None.

2. Final Answer:

   5 <-> 15 <-> 25 <-> None -> Option C

3. Quick Check:

   Two deletes remove last two nodes [OK]

Hint: Each delete removes one tail node [OK]

Common Mistakes:

• Deleting only one node instead of two
• Not updating tail pointer after each delete
• Confusing head and tail nodes