Reverse a Doubly Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap the next and prev pointers of the current node.

DSA Python

current.next, current.prev = current.prev, [1]

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

Acurrent.prev

Bcurrent.next

Chead

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning current.next to current.next instead of current.prev.

Using head or None instead of current.prev.

Swapping current.next and current.prev requires assigning current.prev to current.next and current.next to current.prev. Here, current.next is assigned to current.prev.

2fill in blank

medium

Complete the code to move to the next node in the original list during reversal.

DSA Python

current = [1]

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

Acurrent.prev

Bcurrent.next

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using current.next which now points backward after swapping.

Using head or tail instead of moving current.

After swapping next and prev, the original next node is now current.prev, so we move current to current.prev.

3fill in blank

hard

Fix the error in updating the head pointer after reversal.

DSA Python

if current is None:
    head = [1]

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

Ahead

Bcurrent

Cprev

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head to current which is None.

Assigning head to tail or head itself.

When current becomes None, prev points to the new head of the reversed list, so head should be updated to prev.

4fill in blank

hard

Fill both blanks to complete the reversal loop and update pointers correctly.

DSA Python

while current is not None:
    current.next, current.prev = current.prev, current.next
    prev = [1]
    current = [2]

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

Acurrent

Bprev

Ccurrent.prev

Dcurrent.next

Attempts:

3 left

💡 Hint

Common Mistakes

Updating prev to prev or current.next incorrectly.

Moving current to current.next which is swapped.

Inside the loop, prev is updated to current before moving current to current.prev (the original next node after swapping).

5fill in blank

hard

Fill all three blanks to create a function that reverses a doubly linked list and returns the new head.

DSA Python

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

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

Acurrent

Bcurrent.prev

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using current.next instead of current.prev for moving current.

Not updating head to prev after reversal.

Assigning prev incorrectly inside the loop.

Inside the loop, prev is updated to current, current moves to current.prev (original next), and after loop, head is updated to prev.

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