Delete Node at Specific Position in DSA C# - Time & Space Complexity

Time Complexity: Delete Node at Specific Position

O(n)

Understanding Time Complexity

We want to know how long it takes to remove a node from a linked list at a certain position.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public Node DeleteAtPosition(Node head, int position) {
    if (position == 0) {
        return head.Next;
    }
    Node current = head;
    for (int i = 0; i < position - 1; i++) {
        current = current.Next;
    }
    current.Next = current.Next.Next;
    return head;
}
    

This code removes the node at the given position from a singly linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop moving through nodes to reach the position before the one to delete.
• How many times: Up to position times, depending on where the node is.

How Execution Grows With Input

As the list gets longer, the time to reach the node grows with the position number.

Input Size (n)	Approx. Operations
10	Up to 10 steps to reach the node
100	Up to 100 steps
1000	Up to 1000 steps

Pattern observation: The time grows roughly in direct proportion to the position of the node.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete a node grows linearly with the position in the list.

Common Mistake

[X] Wrong: "Deleting a node anywhere always takes the same constant time."

[OK] Correct: You must first reach the node before the one to delete, which takes longer if the node is far in the list.

Interview Connect

Understanding how linked list operations scale helps you explain your code clearly and shows you know how data size affects performance.

Self-Check

"What if we had a pointer directly to the node before the one to delete? How would the time complexity change?"

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