Delete Node at Specific Position in DSA C# - Build from Scratch

Delete Node at Specific Position

📖 Scenario: You are managing a simple list of tasks represented as a singly linked list. Sometimes, a task needs to be removed from a specific position in the list.

🎯 Goal: Build a program that creates a singly linked list of tasks, then deletes a node at a given position, and finally prints the updated list.

📋 What You'll Learn

Create a singly linked list with exactly these tasks in order: "Task1", "Task2", "Task3", "Task4", "Task5"

Create an integer variable called position and set it to 2

Write a method called DeleteNodeAtPosition that deletes the node at the given position in the linked list

Print the linked list after deletion in the format: Task1 -> Task2 -> Task4 -> Task5 -> null

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality, and managing playlists where items need to be added or removed dynamically.

💼 Career

Understanding linked list operations like deletion is fundamental for software developers working with data structures, improving problem-solving and coding skills.

1

Create the singly linked list

Create a singly linked list class called Node with a string data and a Node reference called next. Then create the linked list with nodes containing these exact strings in order: "Task1", "Task2", "Task3", "Task4", "Task5". Store the head node in a variable called head.

DSA C#

// Define Node class and create linked list with tasks
// Your code here

Hint

Define a Node class with data and next. Then link nodes one by one to form the list.

2

Set the position to delete

Create an integer variable called position and set it to 2 to represent the position of the node to delete (0-based index).

DSA C#

public class Node
{
    public string data;
    public Node next;

    public Node(string data)
    {
        this.data = data;
        this.next = null;
    }
}

Node head = new Node("Task1");
head.next = new Node("Task2");
head.next.next = new Node("Task3");
head.next.next.next = new Node("Task4");
head.next.next.next.next = new Node("Task5");

// Create position variable
// Your code here

Hint

Use int position = 2; to set the position to delete.

3

Write the method to delete node at position

Write a method called DeleteNodeAtPosition that takes Node head and int position as parameters and returns the updated head after deleting the node at the given position. Use 0-based indexing. Handle deleting the head node and other positions correctly.

DSA C#

Node DeleteNodeAtPosition(Node head, int position)
    if (head == null) return head;
    if (position == 0)
        return head.next;
    Node current = head;
    for (int i = 0; current != null && i < position - 1; i++)
        current = current.next;
    if (current == null || current.next == null)
        return head;
    current.next = current.next.next;
    return head;
// Call DeleteNodeAtPosition and print the list

Hint

Check if position is 0 to delete head. Otherwise, move to the node before the position and change its next to skip the node at position.

4

Print the linked list after deletion

Call DeleteNodeAtPosition with head and position, update head with the result, then print the linked list nodes separated by -> and ending with null. The output should be exactly: Task1 -> Task2 -> Task4 -> Task5 -> null

DSA C#

public class Node
{
    public string data;
    public Node next;

    public Node(string data)
    {
        this.data = data;
        this.next = null;
    }
}

Node head = new Node("Task1");
head.next = new Node("Task2");
head.next.next = new Node("Task3");
head.next.next.next = new Node("Task4");
head.next.next.next.next = new Node("Task5");

int position = 2;

Node DeleteNodeAtPosition(Node head, int position)
{
    if (head == null) return head;

    if (position == 0)
    {
        return head.next;
    }

    Node current = head;
    for (int i = 0; current != null && i < position - 1; i++)
    {
        current = current.next;
    }

    if (current == null || current.next == null)
    {
        return head;
    }

    current.next = current.next.next;
    return head;
}

// Call DeleteNodeAtPosition and print the list
// Your code here

Hint

Call DeleteNodeAtPosition with head and position, update head, then loop through the list printing each node's data followed by ->. End with null.

Practice

1. What does deleting a node at a specific position in a singly linked list mean?

easy

A. Adding a new node at that position

B. Removing the node at that position and linking its previous node to the next node

C. Swapping the node at that position with the last node

D. Reversing the entire list up to that position

Solution

1. Step 1: Understand node deletion in linked list

   Deleting a node means removing it and connecting the previous node to the next node to keep the list intact.

2. Step 2: Identify the effect of deletion at a specific position

   The node at the given position is removed, and the previous node's next pointer is updated to skip the deleted node.

3. Final Answer:

   Removing the node at that position and linking its previous node to the next node -> Option B

4. Quick Check:

   Delete node = Remove and link previous to next [OK]

Hint: Delete means remove node and link previous to next [OK]

Common Mistakes:

• Thinking deletion adds a node
• Confusing deletion with swapping nodes
• Assuming deletion reverses the list

2. Which of the following is the correct C# syntax to delete a node at position pos in a singly linked list starting with head?

easy

A. head.next = null;

B. head = head.next[pos];

C. if (pos == 0) head = head.next;

D. head.deleteNode(pos);

Solution

1. Step 1: Check deletion at head position

   If position is 0, deleting means moving head to next node: head = head.next;.

2. Step 2: Validate other options

   head.next = null;: head.next = null; deletes second node regardless of pos. head = head.next[pos];: head = head.next[pos]; invalid syntax (no array indexing on nodes). head.deleteNode(pos);: head.deleteNode(pos); calls non-existent method.

3. Final Answer:

   if (pos == 0) head = head.next; -> Option C

4. Quick Check:

   Delete head = head.next [OK]

Hint: To delete head, assign head to head.next [OK]

Common Mistakes:

• Using array-like indexing on linked list
• Calling undefined methods
• Setting next to null incorrectly

3. What will be the linked list after deleting the node at position 2 from the list: 1 -> 3 -> 5 -> 7 -> null?

medium

A. 1 -> 3 -> 7 -> null

B. 1 -> 5 -> 7 -> null

C. 3 -> 5 -> 7 -> null

D. 1 -> 3 -> 5 -> null

Solution

1. Step 1: Identify node at position 2

   Positions start at 0: 0->1, 1->3, 2->5, 3->7. Node at position 2 is 5.

2. Step 2: Delete node 5 and link previous to next

   Remove 5, link node with value 3 to node with value 7.

3. Final Answer:

   1 -> 3 -> 7 -> null -> Option A

4. Quick Check:

   Delete node 5 at pos 2 = 1->3->7 [OK]

Hint: Count from zero, remove node at position, link neighbors [OK]

Common Mistakes:

• Deleting wrong node by counting from 1
• Not linking previous node to next
• Removing head instead of position 2

4. Given the following code snippet to delete a node at position pos, what is the error?

Node current = head;
for (int i = 0; i < pos - 1; i++) {
    current = current.next;
}
current.next = current.next.next;

medium

A. It does not handle deletion at position 0

B. It causes a syntax error due to missing semicolons

C. It deletes the node after the given position

D. It causes infinite loop

Solution

1. Step 1: Analyze loop for position 0

   For pos=0, loop condition i < pos - 1 (i < -1) is false, current remains head, then current.next = current.next.next; deletes SECOND node (pos 1) instead of head.

2. Step 2: Check deletion logic for other positions

   For pos >=1, loop moves current to node before target and skips target correctly.

3. Final Answer:

   It does not handle deletion at position 0 -> Option A

4. Quick Check:

   Code misses head deletion case [OK]

Hint: Check if code handles deleting head (pos 0) separately [OK]

Common Mistakes:

• Ignoring head deletion case
• Assuming loop runs for pos=0
• Confusing syntax errors with logic errors

5. You have a singly linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to delete the node at position 4. After deletion, what will be the list? Also, what happens if you try to delete at position 5?

hard

A. 10 -> 20 -> 30 -> 40 -> 50 -> null; Deleting at position 5 deletes head

B. 10 -> 20 -> 30 -> 40 -> null; Deleting at position 5 causes no change (position out of range)

C. 10 -> 20 -> 30 -> 40 -> 50 -> null; Deleting at position 5 deletes last node

D. 10 -> 20 -> 30 -> 40 -> null; Deleting at position 5 causes runtime error (null reference)

Solution

1. Step 1: Delete node at position 4

   Positions: 0->10,1->20,2->30,3->40,4->50. Deleting node 50 removes last node, list becomes 10->20->30->40->null.

2. Step 2: Attempt deletion at position 5

   Position 5 does not exist. Trying to access node at pos 5 causes null reference error (runtime error).

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> null; Deleting at position 5 causes runtime error (null reference) -> Option D

4. Quick Check:

   Delete last node works; out-of-range causes error [OK]

Hint: Deleting beyond list length causes runtime error [OK]

Common Mistakes:

• Assuming out-of-range deletion does nothing
• Deleting wrong node at position 5
• Confusing last node position with list length