Delete Node at End in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the last node in the linked list.

DSA C#

while (current.[1] != null) {
    current = current.next;
}

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

Aprev

Btail

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using 'head' or 'tail' which are not node pointers.

We use 'next' to move from one node to the next until we reach the last node where next is null.

2fill in blank

medium

Complete the code to find the second last node before deleting the last node.

DSA C#

while (current.[1].next != null) {
    current = current.next;
}

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

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current.prev.next which is invalid in singly linked lists.

Using current.next.next directly without checking current.next.

We check current.next.next to find the second last node, so current.next must exist.

3fill in blank

hard

Fix the error in setting the last node to null to delete it.

DSA C#

current.[1] = null;

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

Ahead

Bnext

Cprev

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to set 'head' or 'tail' to null inside the node.

Using 'prev' which is not a property in singly linked lists.

To delete the last node, set the 'next' of the second last node to null.

4fill in blank

hard

Fill both blanks to correctly delete the last node from the linked list.

DSA C#

if (head == null || head.[1] == null) {
    head = null;
} else {
    Node current = head;
    while (current.[2].next != null) {
        current = current.next;
    }
    current.next = null;
}

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

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is not available in singly linked lists.

Checking 'tail' which is not a node property.

We check if head.next is null to see if list has one node, then traverse using current.next to find second last node.

5fill in blank

hard

Fill all three blanks to implement a method that deletes the last node and returns the updated head.

DSA C#

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

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

Anext

Bprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is not defined in singly linked lists.

Returning null too early without deleting properly.

We check if head.next is null (one node), traverse using current.next, and set current.next to null to delete last node.

Practice

1. What does the operation Delete Node at End do in a singly linked list?

easy

A. Adds a new node at the end of the list

B. Removes the first node from the list

C. Removes the last node from the list

D. Adds a new node at the beginning of the list

Solution

1. Step 1: Understand the operation meaning

   Deleting a node at the end means removing the last node in the linked list.

2. Step 2: Compare options with operation

   Only Removes the last node from the list matches removing the last node; others add or remove nodes elsewhere.

3. Final Answer:

   Removes the last node from the list -> Option C

4. Quick Check:

   Delete Node at End = Remove last node [OK]

Hint: Deleting at end means removing the last node [OK]

Common Mistakes:

• Confusing delete at end with delete at beginning
• Thinking it adds a node instead of removing
• Assuming it deletes a middle node

2. Which of the following is the correct syntax to delete the last node in a singly linked list in C#?

easy

A. Create a new node and assign it to last node

B. Set head to null directly

C. Set last node's next to head

D. Traverse to second last node, set its next to null

Solution

1. Step 1: Identify correct deletion method

   To delete last node, we must reach second last node and set its next pointer to null.

2. Step 2: Check options for this logic

   Traverse to second last node, set its next to null matches this logic; others do unrelated actions.

3. Final Answer:

   Traverse to second last node, set its next to null -> Option D

4. Quick Check:

   Delete last node = second last node next = null [OK]

Hint: Set second last node's next to null to delete last [OK]

Common Mistakes:

• Setting head to null deletes whole list
• Trying to set last node's next to head causes cycle
• Creating new node does not delete existing node

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> null, what will be the list after deleting the node at the end?

medium

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

B. 10 -> 20 -> 30 -> null

C. 20 -> 30 -> 40 -> null

D. 10 -> 20 -> null

Solution

1. Step 1: Identify last node in list

   The last node is 40 in the list 10 -> 20 -> 30 -> 40 -> null.

2. Step 2: Remove last node and update list

   Deleting last node removes 40, so list becomes 10 -> 20 -> 30 -> null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   Last node 40 removed = 10->20->30->null [OK]

Hint: Remove last value to get new list [OK]

Common Mistakes:

• Not removing last node, list unchanged
• Removing first node instead
• Removing more than one node accidentally

4. Consider this code snippet to delete the last node of a singly linked list:

Node current = head;
while(current.next.next != null) {
    current = current.next;
}
current.next = null;

What error can occur when this code runs on a list with only one node?

medium

A. NullReferenceException because current.next.next is invalid

B. No error, code works fine

C. Infinite loop because condition never false

D. Deletes the head node incorrectly

Solution

1. Step 1: Analyze condition for single node list

   If list has one node, current.next is null, so current.next.next causes error.

2. Step 2: Identify exception type

   Accessing next of null causes NullReferenceException in C#.

3. Final Answer:

   NullReferenceException because current.next.next is invalid -> Option A

4. Quick Check:

   Single node list causes null.next access error [OK]

Hint: Check for null before accessing next.next [OK]

Common Mistakes:

• Ignoring null check for single node
• Assuming loop runs safely always
• Thinking it deletes head node

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> 45 -> null. Write a method to delete the last node and return the updated list. Which of the following code snippets correctly implements this?

hard

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

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

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

D. if (head == null) return null;
head = head.next;
return head;

Solution

1. Step 1: Handle empty and single node cases

   If list is empty or has one node, deleting last node returns null (empty list).

2. Step 2: Traverse to second last node and delete last

   Loop until current.next.next is null, then set current.next to null to remove last node.

3. Step 3: Return updated head

   Return head pointer to updated list.

4. Final Answer:

   Code snippet D correctly deletes last node and returns updated list -> Option A

5. Quick Check:

   Check null, traverse to second last, set next null [OK]

Hint: Check list size, then remove last by second last next = null [OK]

Common Mistakes:

• Setting current to null does not remove last node
• Assigning current.next to current creates cycle
• Deleting head instead of last node