Reverse a Doubly Linked List in DSA Python - Time & Space Complexity

Time Complexity: Reverse a Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a doubly linked list changes as the list grows.

Specifically, how does the number of steps grow when the list gets longer?

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

def reverse_doubly_linked_list(head):
    current = head
    while current:
        current.prev, current.next = current.next, current.prev
        head = current
        current = current.prev
    return head

This code reverses the links of a doubly linked list so the last node becomes the first.

Identify Repeating Operations

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node, so n times for n nodes.

How Execution Grows With Input

As the list grows, the number of steps grows directly with the number of nodes.

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

Pattern observation: The steps increase in a straight line as the list gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows directly with the number of nodes.

Common Mistake

[X] Wrong: "Reversing a doubly linked list takes more than linear time because of the two pointers per node."

[OK] Correct: Each node is visited once, and swapping two pointers is a constant time action, so overall time is still linear.

Interview Connect

Understanding this helps you explain how linked list operations scale, a common topic in interviews.

Self-Check

"What if we tried to reverse the list using recursion instead of a loop? How would the time complexity change?"

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To change the direction of traversal so the last node becomes the first

B. To delete all nodes from the list

C. To sort the nodes in ascending order

D. To convert it into a singly linked list

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes with pointers to both previous and next nodes, allowing traversal in both directions.

2. Step 2: Purpose of reversing

   Reversing swaps the previous and next pointers of each node, making the last node the new head and changing traversal direction.

3. Final Answer:

   To change the direction of traversal so the last node becomes the first -> Option A

4. Quick Check:

   Reversing swaps pointers to reverse traversal [OK]

Hint: Reversing swaps previous and next pointers [OK]

Common Mistakes:

• Thinking reversing deletes nodes
• Confusing reversing with sorting
• Assuming it converts to singly linked list

2. Which of the following is the correct way to swap the next and prev pointers of a node curr in Python?

easy

A. curr.next = curr.prev; curr.prev = curr.next

B. curr.next, curr.prev = curr.prev, curr.next

C. curr.prev = curr.next; curr.next = curr.prev

D. curr.next = curr.next; curr.prev = curr.prev

Solution

1. Step 1: Understand Python tuple unpacking

   Using a, b = b, a swaps values of a and b simultaneously without temporary variable.

2. Step 2: Apply to pointers

   Swapping curr.next and curr.prev correctly uses curr.next, curr.prev = curr.prev, curr.next.

3. Final Answer:

   curr.next, curr.prev = curr.prev, curr.next -> Option B

4. Quick Check:

   Tuple unpacking swaps pointers correctly [OK]

Hint: Use tuple unpacking to swap pointers in one line [OK]

Common Mistakes:

• Assigning sequentially causing both to be same
• Not swapping both pointers
• Using incorrect order in assignment

3. Given the doubly linked list: 1 <-> 2 <-> 3 <-> null, what will be the list after reversing it using the standard reverse algorithm?

medium

A. 3 <-> 2 <-> 1 <-> null

B. 1 <-> 2 <-> 3 <-> null

C. null <-> 3 <-> 2 <-> 1

D. 3 -> 2 -> 1 -> null (singly linked list)

Solution

1. Step 1: Original list traversal

   The list is 1 <-> 2 <-> 3 <-> null, head at 1, tail at 3.

2. Step 2: Reverse pointers

   Swapping next and prev pointers reverses the list, making 3 the new head and 1 the new tail.

3. Final Answer:

   3 <-> 2 <-> 1 <-> null -> Option A

4. Quick Check:

   Reversed list order is 3, 2, 1 [OK]

Hint: Reversed list order is original tail to head [OK]

Common Mistakes:

• Confusing original order with reversed
• Misplacing null pointer
• Assuming singly linked list after reverse

4. What is the error in this code snippet for reversing a doubly linked list?

def reverse(head):
    curr = head
    while curr:
        curr.next = curr.prev
        curr.prev = curr.next
        curr = curr.prev
    return head

medium

A. The code correctly reverses the list

B. The loop condition should be while curr.next

C. Pointers are not swapped simultaneously causing incorrect links

D. The function should return curr instead of head

Solution

1. Step 1: Analyze pointer swapping

   Assigning curr.next = curr.prev then curr.prev = curr.next sequentially overwrites curr.next before using it, losing original value.

2. Step 2: Effect on traversal

   Because pointers are not swapped simultaneously, traversal using curr = curr.prev fails and links become incorrect.

3. Final Answer:

   Pointers are not swapped simultaneously causing incorrect links -> Option C

4. Quick Check:

   Swap pointers simultaneously to avoid overwriting [OK]

Hint: Swap pointers in one line to avoid overwriting [OK]

Common Mistakes:

• Swapping pointers in two separate assignments
• Returning old head instead of new head
• Incorrect loop condition

5. You have a doubly linked list with nodes containing integer values. After reversing it, you want to print the list from the new head to the end. Which of the following code snippets correctly prints the reversed list?

hard

A. curr = reversed_head
while curr:
print(curr.data, end=' <-> ')
curr = curr.prev
print('null')

B. curr = reversed_head
while curr.prev:
print(curr.data, end=' <-> ')
curr = curr.prev
print('null')

C. curr = original_head
while curr:
print(curr.data, end=' <-> ')
curr = curr.next
print('null')

D. curr = reversed_head
while curr:
print(curr.data, end=' <-> ')
curr = curr.next
print('null')

Solution

1. Step 1: Identify traversal direction after reversing

   After reversing, the new head points to the start of the reversed list, and traversal should use next pointers to move forward.

2. Step 2: Correct printing loop

   Starting from reversed_head, printing curr.data and moving with curr = curr.next prints the list in correct order until curr is null.

3. Final Answer:

   curr = reversed_head while curr: print(curr.data, end=' <-> ') curr = curr.next print('null') -> Option D

4. Quick Check:

   Traverse reversed list using next pointers from new head [OK]

Hint: Print using next pointers from new head after reverse [OK]

Common Mistakes:

• Using prev pointer to traverse forward
• Starting from original head after reverse
• Loop condition stopping early