Delete from End of Doubly Linked List in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the doubly linked list is empty before deletion.

DSA C#

if (head == [1]) {
    return;
}

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

Anull

Bhead

Ctail

Dthis

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if head equals tail instead of null.

Using 'this' keyword incorrectly.

We check if head is null to know if the list is empty before deleting.

2fill in blank

medium

Complete the code to move the tail pointer to the previous node before deletion.

DSA C#

tail = tail.[1];

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

Aprev

Bparent

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'next' instead of 'prev'.

Trying to assign tail to head.

To delete from the end, we move tail to its previous node.

3fill in blank

hard

Fix the error in setting the new tail's next pointer to null after deletion.

DSA C#

if (tail != null) {
    tail.[1] = null;
}

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

Ahead

Bprev

Cnext

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Setting tail.prev to null instead of tail.next.

Setting tail to null.

After moving tail back, we set tail.next to null to mark the end.

4fill in blank

hard

Fill both blanks to correctly handle the case when the list has only one node.

DSA C#

if (head == tail) {
    head = [1];
    tail = [2];
}

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

Anull

Bhead

Ctail

Dthis

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head or tail to themselves instead of null.

Not handling the single-node case.

When only one node exists, deleting it sets both head and tail to null.

5fill in blank

hard

Fill all three blanks to complete the DeleteFromEnd method of a doubly linked list.

DSA C#

public void DeleteFromEnd() {
    if (head == [1]) {
        return;
    }
    if (head == tail) {
        head = [2];
        tail = [3];
        return;
    }
    tail = tail.prev;
    tail.next = null;
}

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

Anull

Bhead

Ctail

Dthis

Attempts:

3 left

💡 Hint

Common Mistakes

Not returning after setting head and tail to null.

Incorrectly checking for empty list.

Not updating tail.next to null.

The method checks if the list is empty, then if it has one node, sets head and tail to null, else moves tail back and disconnects the last node.

Practice

1. What happens when you delete the last node from a doubly linked list that has multiple nodes?

easy

A. The first node is deleted instead.

B. The head pointer is set to null.

C. The list becomes empty immediately.

D. The second last node becomes the new tail and its next pointer is set to null.

Solution

1. Step 1: Identify the last node and its previous node

   In a doubly linked list, the last node's previous node is the second last node.

2. Step 2: Update the second last node's next pointer

   Set the second last node's next pointer to null to remove the last node from the list.

3. Final Answer:

   The second last node becomes the new tail and its next pointer is set to null. -> Option D

4. Quick Check:

   Deleting last node updates tail's next to null [OK]

Hint: Last node's previous next pointer must be null after deletion [OK]

Common Mistakes:

• Setting head to null instead of updating tail
• Not updating the previous node's next pointer
• Deleting the first node by mistake

2. Which of the following is the correct C# syntax to delete the last node from a doubly linked list named dll?

easy

A. dll.Head = dll.Head.Next; dll.Head.Previous = null;

B. dll.Tail = dll.Tail.Previous; dll.Tail.Next = null;

C. dll.Tail = null; dll.Head = null;

D. dll.Head.Previous = dll.Tail; dll.Tail.Next = dll.Head;

Solution

1. Step 1: Move tail pointer to previous node

   Set dll.Tail to dll.Tail.Previous to remove the last node.

2. Step 2: Set new tail's next pointer to null

   Update dll.Tail.Next to null to mark the end of the list.

3. Final Answer:

   dll.Tail = dll.Tail.Previous; dll.Tail.Next = null; -> Option B

4. Quick Check:

   Correct tail update syntax = dll.Tail = dll.Tail.Previous; dll.Tail.Next = null; [OK]

Hint: Tail moves back, next pointer cleared [OK]

Common Mistakes:

• Updating head instead of tail
• Not setting tail's next to null
• Setting both head and tail to null unnecessarily

3. Consider the following C# code that deletes the last node from a doubly linked list. What will be the printed list after deletion?

dll.AddLast(1); dll.AddLast(2); dll.AddLast(3); dll.DeleteFromEnd(); dll.PrintList();

medium

A. 2 -> 3 -> null

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

C. 1 -> 2 -> null

D. null

Solution

1. Step 1: Add nodes 1, 2, 3 to the list

   The list becomes 1 -> 2 -> 3 -> null.

2. Step 2: Delete the last node (3)

   After deletion, the list becomes 1 -> 2 -> null.

3. Final Answer:

   1 -> 2 -> null -> Option C

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Deleting last removes tail node only [OK]

Common Mistakes:

• Assuming list remains unchanged
• Deleting head instead of tail
• Printing list before deletion

4. The following code attempts to delete the last node from a doubly linked list but causes a runtime error. What is the error?

if (dll.Tail != null) { dll.Tail = dll.Tail.Previous; dll.Tail.Next = null; }

medium

A. NullReferenceException when list has only one node

B. Syntax error due to missing braces

C. The head pointer is not updated

D. The list becomes circular

Solution

1. Step 1: Check deletion when only one node exists

   If only one node exists, dll.Tail.Previous is null, so dll.Tail becomes null.

2. Step 2: Accessing dll.Tail.Next causes error

   After setting dll.Tail to null, accessing dll.Tail.Next throws NullReferenceException.

3. Final Answer:

   NullReferenceException when list has only one node -> Option A

4. Quick Check:

   Accessing property on null tail causes error [OK]

Hint: Check for null tail before accessing Next [OK]

Common Mistakes:

• Not handling single-node list case
• Ignoring null checks before pointer access
• Assuming head updates automatically

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 40. After deleting the last node twice, what will be the list and the tail node's value?

hard

A. 10 <-> 20 -> null, Tail = 20

B. 10 <-> 20 <-> 30 -> null, Tail = 30

C. 10 -> null, Tail = 10

D. Empty list, Tail = null

Solution

1. Step 1: Initial list and first deletion

   List is 10 <-> 20 <-> 30 <-> 40. Deleting last node removes 40, list becomes 10 <-> 20 <-> 30.

2. Step 2: Second deletion removes 30

   After second deletion, list becomes 10 <-> 20, and tail points to node with value 20.

3. Final Answer:

   10 <-> 20 -> null, Tail = 20 -> Option A

4. Quick Check:

   Two deletions remove last two nodes [OK]

Hint: Each deletion removes tail, tail moves back [OK]

Common Mistakes:

• Deleting only one node instead of two
• Not updating tail after deletions
• Confusing head and tail pointers