Delete Node at Specific Position in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting node at position 2 in a singly linked list

What is the printed state of the linked list after deleting the node at position 2?

DSA C#

using System;

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

class LinkedList {
    public Node head;

    public void DeleteNodeAtPosition(int position) {
        if (head == null) return;
        Node temp = head;
        if (position == 0) {
            head = temp.next;
            return;
        }
        for (int i = 0; temp != null && i < position - 1; i++) {
            temp = temp.next;
        }
        if (temp == null || temp.next == null) return;
        temp.next = temp.next.next;
    }

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

class Program {
    static void Main() {
        LinkedList list = new LinkedList();
        list.head = new Node(10);
        list.head.next = new Node(20);
        list.head.next.next = new Node(30);
        list.head.next.next.next = new Node(40);
        list.DeleteNodeAtPosition(2);
        list.PrintList();
    }
}

A 10 -> 20 -> 40 -> null

B 10 -> 30 -> 40 -> null

C 20 -> 30 -> 40 -> null

D 10 -> 20 -> 30 -> null

💡 Hint

Remember that position 0 is the head node. Deleting at position 2 removes the third node.

The node at position 2 (0-based) is the node with value 30. After deletion, the list links 20 directly to 40.

❓ Predict Output

intermediate

Output after deleting node at position 0 (head) in a singly linked list

What is the printed state of the linked list after deleting the node at position 0?

DSA C#

using System;

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

class LinkedList {
    public Node head;

    public void DeleteNodeAtPosition(int position) {
        if (head == null) return;
        Node temp = head;
        if (position == 0) {
            head = temp.next;
            return;
        }
        for (int i = 0; temp != null && i < position - 1; i++) {
            temp = temp.next;
        }
        if (temp == null || temp.next == null) return;
        temp.next = temp.next.next;
    }

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

class Program {
    static void Main() {
        LinkedList list = new LinkedList();
        list.head = new Node(5);
        list.head.next = new Node(15);
        list.head.next.next = new Node(25);
        list.DeleteNodeAtPosition(0);
        list.PrintList();
    }
}

A 5 -> 25 -> null

B 15 -> 25 -> null

C 5 -> 15 -> 25 -> null

D null

💡 Hint

Deleting at position 0 removes the head node.

The head node with value 5 is removed, so the new head is the node with value 15.

🔧 Debug

advanced

Identify the error when deleting a node at an invalid position

What error or behavior occurs when trying to delete a node at position 5 in a list of 3 nodes?

DSA C#

using System;

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

class LinkedList {
    public Node head;

    public void DeleteNodeAtPosition(int position) {
        if (head == null) return;
        Node temp = head;
        if (position == 0) {
            head = temp.next;
            return;
        }
        for (int i = 0; temp != null && i < position - 1; i++) {
            temp = temp.next;
        }
        if (temp == null || temp.next == null) return;
        temp.next = temp.next.next;
    }
}

class Program {
    static void Main() {
        LinkedList list = new LinkedList();
        list.head = new Node(1);
        list.head.next = new Node(2);
        list.head.next.next = new Node(3);
        list.DeleteNodeAtPosition(5);
        Console.WriteLine("Deletion attempted at position 5");
    }
}

A NullReferenceException at runtime

B Compilation error due to invalid loop

C IndexOutOfRangeException at runtime

D No error, list remains unchanged

💡 Hint

Check the conditions that stop deletion if position is invalid.

The code checks if temp or temp.next is null before deletion. If position is out of range, it returns without error.

❓ Predict Output

advanced

Output after deleting last node in a singly linked list

What is the printed state of the linked list after deleting the node at the last position?

DSA C#

using System;

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

class LinkedList {
    public Node head;

    public void DeleteNodeAtPosition(int position) {
        if (head == null) return;
        Node temp = head;
        if (position == 0) {
            head = temp.next;
            return;
        }
        for (int i = 0; temp != null && i < position - 1; i++) {
            temp = temp.next;
        }
        if (temp == null || temp.next == null) return;
        temp.next = temp.next.next;
    }

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

class Program {
    static void Main() {
        LinkedList list = new LinkedList();
        list.head = new Node(7);
        list.head.next = new Node(14);
        list.head.next.next = new Node(21);
        list.head.next.next.next = new Node(28);
        list.DeleteNodeAtPosition(3);
        list.PrintList();
    }
}

A 7 -> 14 -> 28 -> null

B 7 -> 14 -> 21 -> 28 -> null

C 7 -> 14 -> 21 -> null

D 14 -> 21 -> 28 -> null

💡 Hint

Deleting at position 3 removes the last node in a 4-node list.

The node with value 28 at position 3 is removed, so the list ends at 21.

🧠 Conceptual

expert

Effect of deleting a node at a position greater than list length

If you try to delete a node at a position greater than the length of the singly linked list, what is the expected behavior of the given DeleteNodeAtPosition method?

DSA C#

public void DeleteNodeAtPosition(int position) {
    if (head == null) return;
    Node temp = head;
    if (position == 0) {
        head = temp.next;
        return;
    }
    for (int i = 0; temp != null && i < position - 1; i++) {
        temp = temp.next;
    }
    if (temp == null || temp.next == null) return;
    temp.next = temp.next.next;
}

A The method does nothing and the list remains unchanged

B The method deletes the last node in the list

C The method throws a NullReferenceException at runtime

D The method deletes the head node

💡 Hint

Look at the checks for temp and temp.next before deletion.

If position is out of range, temp or temp.next becomes null and the method returns without deleting any node.

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