Reverse a Doubly Linked List in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

current.[1] = temp;

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

Aprev

Bnext

Cdata

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping data instead of pointers

Assigning to current.next instead of current.prev

We swap the next and prev pointers. Here, current.prev is assigned temp which holds the original next node.

2fill in blank

medium

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

DSA C#

current = current.[1];

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

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using current.next which now points backwards

Using head or tail instead of pointers

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

3fill in blank

hard

Fix the error in updating the head pointer after the reversal loop.

DSA C#

if (temp != null) head = temp.[1];

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

Anext

Bdata

Cprev

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head to temp.next which is null

Setting head to temp.data instead of a node pointer

After the loop, temp points to the last processed node, so head should be updated to temp.prev which is the new head.

4fill in blank

hard

Fill both blanks to correctly swap pointers and move current in the reversal loop.

DSA C#

Node temp = current.[1];
current.[2] = current.[1];
current.[1] = temp;
current = current.[2];

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

Anext

Bprev

Cdata

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up next and prev assignments

Not moving current correctly after swapping

We first save current.next in temp, then assign current.next to current.prev, then current.prev to temp, and finally move current to current.prev (which is saved in current.next before swap).

5fill in blank

hard

Fill all three blanks to complete the reversal function for a doubly linked list.

DSA C#

public Node Reverse(Node head) {
    Node current = head;
    Node temp = null;
    while (current != null) {
        temp = current.[1];
        current.[2] = current.[3];
        current.[3] = temp;
        current = current.[2];
    }
    if (temp != null) head = temp.[1];
    return head;
}

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

Anext

Bprev

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing next and prev in assignments

Not updating head correctly after reversal

Inside the loop, temp saves current.next, then current.prev is assigned current.next, current.next is assigned temp, and current moves to current.prev. After the loop, head is updated to temp.prev which is the new head.

Practice

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

easy

A. To change the order of nodes so the last becomes first and vice versa

B. To delete all nodes from the list

C. To add new nodes at the end of the list

D. To sort the nodes in ascending order

Solution

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

   A doubly linked list has nodes connected forward and backward using next and prev pointers.

2. Step 2: Purpose of reversing

   Reversing swaps the direction so the last node becomes the first and the first becomes the last.

3. Final Answer:

   To change the order of nodes so the last becomes first and vice versa -> Option A

4. Quick Check:

   Reversing = changing node order [OK]

Hint: Reversing flips the list order front to back [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Mixing up adding nodes with reversing

2. Which of the following C# code snippets correctly swaps the next and prev pointers of a node in a doubly linked list?

easy

A. node.next = node.prev; node.prev = node.next;

B. node.prev = node.next; node.next = node.prev;

C. var temp = node.prev; node.next = node.prev; node.prev = temp;

D. var temp = node.next; node.next = node.prev; node.prev = temp;

Solution

1. Step 1: Understand swapping pointers

   To swap two pointers, use a temporary variable to hold one value before overwriting it.

2. Step 2: Check each option

   The correct snippet saves node.next in temp, assigns node.next = node.prev, then node.prev = temp. Direct assignments without temp cause data loss. The flawed temp usage assigns node.next = node.prev without saving the original node.next first, losing data.

3. Final Answer:

   var temp = node.next; node.next = node.prev; node.prev = temp; -> Option D

4. Quick Check:

   Use temp variable to swap pointers [OK]

Hint: Always use a temp variable to swap two pointers [OK]

Common Mistakes:

• Swapping without a temporary variable
• Assigning pointers in wrong order
• Confusing next and prev roles

3. What will be the output of the following C# code after reversing the doubly linked list?

class Node {
  public int data;
  public Node next, prev;
  public Node(int d) { data = d; next = prev = null; }
}

// List: 1 <-> 2 <-> 3
Node head = new Node(1);
Node second = new Node(2);
Node third = new Node(3);
head.next = second; second.prev = head;
second.next = third; third.prev = second;

// Reverse function
Node Reverse(Node head) {
  Node current = head;
  Node temp = null;
  while (current != null) {
    temp = current.prev;
    current.prev = current.next;
    current.next = temp;
    current = current.prev;
  }
  if (temp != null) head = temp.prev;
  return head;
}

head = Reverse(head);

// Print list forward
Node tempNode = head;
string result = "";
while (tempNode != null) {
  result += tempNode.data + " -> ";
  tempNode = tempNode.next;
}
result += "null";
Console.WriteLine(result);

medium

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

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

C. null

D. 1 -> 3 -> 2 -> null

Solution

1. Step 1: Trace the reversal loop

   Each node swaps its next and prev pointers. The loop moves current to current.prev (which was originally next). After the loop, head is updated to the last processed node.

2. Step 2: Verify the new order

   Original list: 1 <-> 2 <-> 3. After reversal, the list becomes 3 <-> 2 <-> 1. Printing forward from new head prints 3 -> 2 -> 1 -> null.

3. Final Answer:

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

4. Quick Check:

   Reversed list prints backward order [OK]

Hint: Reversing flips list order; last node becomes head [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing next and prev pointers
• Printing list without updating head

4. Identify the bug in this reverse function for a doubly linked list:

Node Reverse(Node head) {
  Node current = head;
  Node temp = null;
  while (current != null) {
    temp = current.next;
    current.next = current.prev;
    current.prev = temp;
    current = current.next;
  }
  if (temp != null) head = temp.prev;
  return head;
}

medium

A. The variable temp should store current.next, not current.prev

B. The loop should move current to current.prev, not current.next

C. The head update line should be head = temp.next

D. There is no bug; the code is correct

Solution

1. Step 1: Analyze pointer swapping

   The code swaps next and prev pointers correctly, but moves current to current.next, which after swapping points backward.

2. Step 2: Correct traversal direction

   After swapping, current should move to current.prev (originally next) to continue forward traversal. Moving to current.next causes skipping or infinite loop.

3. Final Answer:

   The loop should move current to current.prev, not current.next -> Option B

4. Quick Check:

   Traverse using swapped pointers correctly [OK]

Hint: After swapping, move current to current.prev pointer [OK]

Common Mistakes:

• Moving current in wrong direction after swap
• Incorrect temp assignment
• Wrong head update after reversal

5. Given a doubly linked list with nodes containing integer values, how would you modify the reversal function to also count how many nodes were reversed during the process?

hard

A. Add a counter but increment it only after the loop finishes.

B. Count nodes before reversal and return the count without changes.

C. Add a counter variable initialized to 0, increment it inside the reversal loop, and return it along with the new head.

D. Use recursion to reverse and count nodes simultaneously.

Solution

1. Step 1: Initialize a counter before the loop

   Start a counter at zero to track how many nodes are processed during reversal.

2. Step 2: Increment counter inside the reversal loop

   Each time a node is processed (pointers swapped), increase the counter by one.

3. Step 3: Return both new head and count

   Modify the function to return the new head and the count of reversed nodes.

4. Final Answer:

   Add a counter variable initialized to 0, increment it inside the reversal loop, and return it along with the new head. -> Option C

5. Quick Check:

   Count nodes during reversal loop [OK]

Hint: Count nodes inside reversal loop for accurate total [OK]

Common Mistakes:

• Counting nodes before reversal only
• Incrementing counter after loop ends
• Not returning the count with the head