Delete from End of Doubly Linked List in DSA Python - Time & Space Complexity

Time Complexity: Delete from End of Doubly Linked List

O(1)

Understanding Time Complexity

When we delete a node from the end of a doubly linked list, we want to know how the time it takes changes as the list grows.

We ask: How many steps does it take to remove the last node when the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

class DoublyLinkedList:
    def __init__(self):
        self.head = None
        self.tail = None

    def delete_from_end(self):
        if not self.tail:
            return  # List is empty
        if self.head == self.tail:
            self.head = None
            self.tail = None
        else:
            self.tail = self.tail.prev
            self.tail.next = None
    

This code removes the last node from a doubly linked list by updating pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing the tail node and updating pointers.
• How many times: This happens once per deletion, no loops or recursion involved.

How Execution Grows With Input

Deleting from the end always takes the same number of steps, no matter how big the list is.

Input Size (n)	Approx. Operations
10	5 steps
100	5 steps
1000	5 steps

Pattern observation: The number of steps stays the same as the list grows.

Final Time Complexity

Time Complexity: O(1)

This means deleting from the end takes a constant amount of time regardless of list size.

Common Mistake

[X] Wrong: "Deleting from the end takes longer as the list grows because we have to find the last node."

[OK] Correct: The doubly linked list keeps a direct reference to the last node (tail), so no searching is needed.

Interview Connect

Knowing that deleting from the end of a doubly linked list is quick shows you understand how pointers help save time in data structures.

Self-Check

"What if we did not keep a tail pointer? How would the time complexity of deleting from the end change?"

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