Delete Node at Specific Position in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the first step when deleting a node at a specific position in a singly linked list?

Check if the list is empty or if the position is valid (e.g., position is not negative and within the list length).

beginner

How do you delete the head node (position 0) in a singly linked list?

Set the head pointer to the next node of the current head, effectively removing the first node from the list.

intermediate

Why do you need to keep track of the previous node when deleting a node at a specific position (other than head)?

Because you need to update the previous node's next pointer to skip the node being deleted and point to the node after it.

beginner

What happens if the position to delete is greater than the length of the linked list?

No deletion occurs because the position is invalid; the list remains unchanged.

intermediate

In C#, what is the effect of setting a node's next pointer to null during deletion?

It helps the garbage collector to reclaim the memory of the removed node since no references point to it anymore.

What should you do first before deleting a node at a specific position in a linked list?

A Check if the list is empty or position is valid

B Delete the head node directly

C Traverse to the last node

D Create a new node

Always check if the list is empty or if the position is valid before attempting deletion.

How do you delete the node at position 0 in a singly linked list?

A Set head to head.next

B Set head to null

C Traverse to the last node

D Set head.next to null

Deleting the head means moving the head pointer to the next node.

When deleting a node at position > 0, what pointer do you update?

A Tail pointer

B Head pointer

C Deleted node's next pointer

D Previous node's next pointer

You update the previous node's next pointer to skip the deleted node.

What happens if the position to delete is outside the list length?

A Deletes the last node

B No deletion occurs

C Deletes the head node

D Throws an exception

Invalid positions do not cause deletion; the list stays the same.

Why is setting the deleted node's next pointer to null helpful in C#?

A Improves performance

B Prevents memory leaks manually

C Helps garbage collection

D Is required to delete the node

Removing references allows the garbage collector to reclaim memory.

Explain the steps to delete a node at a specific position in a singly linked list.

Think about how pointers are updated to remove the node.

Describe how deleting the head node differs from deleting a node at any other position in a singly linked list.

Focus on pointer updates and traversal differences.

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