Reverse a Doubly Linked List in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Reversing a Doubly Linked List

What is the printed state of the doubly linked list after reversing it using the given method?

DSA C#

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

public class DoublyLinkedList {
    public Node head;

    public void Reverse() {
        Node temp = null;
        Node current = head;

        while (current != null) {
            temp = current.prev;
            current.prev = current.next;
            current.next = temp;
            current = current.prev;
        }

        if (temp != null) {
            head = temp.prev;
        }
    }

    public void PrintList() {
        Node node = head;
        while (node != null) {
            System.Console.Write(node.data + " <-> ");
            node = node.next;
        }
        System.Console.WriteLine("null");
    }
}

// Setup
DoublyLinkedList dll = new DoublyLinkedList();
dll.head = new Node(1);
Node second = new Node(2);
Node third = new Node(3);

// Linking nodes

dll.head.next = second;
second.prev = dll.head;
second.next = third;
third.prev = second;

// Reverse and print

dll.Reverse();
dll.PrintList();

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

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

C null

D 3 <-> 2 <-> 1 <-> null

💡 Hint

Think about how the prev and next pointers are swapped for each node.

The Reverse method swaps the prev and next pointers for each node, then updates the head to the last node processed. This reverses the list order from 1->2->3 to 3->2->1.

🧠 Conceptual

intermediate

Understanding Pointer Swapping in Reverse

In the reverse method of a doubly linked list, what is the main purpose of swapping the prev and next pointers of each node?

A To delete nodes from the list

B To reverse the direction of the list by changing the links between nodes

C To sort the list in ascending order

D To create a new list with copied nodes

💡 Hint

Think about what happens to the direction of links when prev and next are swapped.

Swapping prev and next pointers reverses the direction of the links between nodes, effectively reversing the list.

🔧 Debug

advanced

Identify the Bug in Reverse Method

What error will occur if the line 'head = temp.prev;' is replaced with 'head = temp.next;' in the reverse method?

DSA C#

public void Reverse() {
    Node temp = null;
    Node current = head;

    while (current != null) {
        temp = current.prev;
        current.prev = current.next;
        current.next = temp;
        current = current.prev;
    }

    if (temp != null) {
        head = temp.next; // changed from temp.prev
    }
}

A The list will be partially reversed with some nodes lost

B The list will be reversed correctly

C A runtime NullReferenceException will occur

D The head will point to null, resulting in an empty list

💡 Hint

Consider what temp points to at the end of the loop and what temp.next would be.

At the end of the loop, temp points to the previous node of the last processed node. Using temp.next sets head to an intermediate node in the reversed list, resulting in only part of the list being accessible from head and some nodes lost.

❓ Predict Output

advanced

Output After Reversing a Longer Doubly Linked List

Given the doubly linked list 10 <-> 20 <-> 30 <-> 40 <-> null, what is the printed output after reversing it with the standard reverse method?

DSA C#

DoublyLinkedList dll = new DoublyLinkedList();
dll.head = new Node(10);
Node n2 = new Node(20);
Node n3 = new Node(30);
Node n4 = new Node(40);

dll.head.next = n2; n2.prev = dll.head;
n2.next = n3; n3.prev = n2;
n3.next = n4; n4.prev = n3;

dll.Reverse();
dll.PrintList();

A 40 <-> 30 <-> 20 <-> 10 <-> null

B 10 <-> 20 <-> 30 <-> 40 <-> null

C null

D 40 <-> 20 <-> 30 <-> 10 <-> null

💡 Hint

The reverse method swaps prev and next pointers for all nodes.

Reversing swaps the links so the list order is reversed from 10->20->30->40 to 40->30->20->10.

🧠 Conceptual

expert

Why is Updating Head Necessary After Reversing?

After swapping prev and next pointers for all nodes in a doubly linked list, why must we update the head pointer to the last processed node's previous pointer?

A Because the head pointer should point to the node with the largest value after reversal

B Because the head pointer must always point to the node with the smallest value

C Because the original head becomes the tail, and the new head is the last node processed, which is accessed via temp.prev

D Because the head pointer must be reset to null to mark the end of the list

💡 Hint

Think about the position of the head before and after reversal.

After reversal, the original head is at the end (tail), so the head pointer must be updated to the last node processed, which is temp.prev, to correctly point to the new start of the list.

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