Delete Node at Specific Position in DSA Javascript - 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 list at a certain spot.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function deleteNode(head, position) {
  if (position === 0) {
    return head.next;
  }
  let current = head;
  for (let 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

• Primary operation: The for-loop that moves through the list nodes.
• How many times: It runs up to position - 1 times to reach the node before the one to delete.

How Execution Grows With Input

As the list gets longer, the time to reach the position grows roughly in a straight line.

Input Size (n)	Approx. Operations
10	Up to 9 steps
100	Up to 99 steps
1000	Up to 999 steps

Pattern observation: The steps increase directly with the position number.

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 at any position always takes the same time."

[OK] Correct: Because you must first reach the node before the one to delete, which takes longer if the position is farther in the list.

Interview Connect

Understanding how list traversal affects time helps you explain and write efficient code for linked list operations.

Self-Check

"What if we had a doubly linked list? How would the time complexity change when deleting a node at a specific position?"

Practice

1. What happens when you delete a node at position 0 in a singly linked list?

easy

A. The head pointer moves to the next node, effectively removing the first node.

B. The last node is removed from the list.

C. The list becomes empty regardless of its size.

D. The node at position 1 is deleted instead.

Solution

1. Step 1: Understand position 0 in linked list

   Position 0 refers to the first node, also called the head.

2. Step 2: Effect of deleting head node

   Deleting the head means moving the head pointer to the next node, removing the first node from the list.

3. Final Answer:

   The head pointer moves to the next node, effectively removing the first node. -> Option A

4. Quick Check:

   Deleting position 0 removes the first node [OK]

Hint: Deleting position 0 means changing the head pointer [OK]

Common Mistakes:

• Thinking the last node is deleted instead
• Assuming the list becomes empty always
• Deleting the wrong node position

2. Which of the following is the correct syntax to delete a node at position pos in a singly linked list in JavaScript?

easy

A. head = head[pos + 1];

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

C. delete head[pos];

D. head.next = null;

Solution

1. Step 1: Check deletion at position 0

   Deleting at position 0 means moving head to head.next.

2. Step 2: Validate syntax correctness

   if (pos === 0) head = head.next; correctly assigns head to head.next. Other options misuse syntax or do not delete the node properly.

3. Final Answer:

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

4. Quick Check:

   Correct syntax for deleting head node is head = head.next [OK]

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

Common Mistakes:

• Using delete keyword incorrectly on linked list nodes
• Assigning head.next to null without updating head
• Accessing nodes with array-like syntax

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> null, what is the list after deleting the node at position 2?

medium

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

B. 10 -> 30 -> 40 -> null

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

D. 10 -> 20 -> 40 -> null

Solution

1. Step 1: Identify node at position 2

   Positions start at 0, so position 2 is the third node with value 30.

2. Step 2: Delete node at position 2

   Remove node 30 by linking node 20's next to node 40.

3. Final Answer:

   10 -> 20 -> 40 -> null -> Option D

4. Quick Check:

   Deleted node 30 at position 2 [OK]

Hint: Count from 0, remove node at that index [OK]

Common Mistakes:

• Deleting wrong node by miscounting positions
• Not updating the previous node's next pointer
• Confusing node values with positions

4. What is the error in the following code snippet for deleting a node at position pos in a singly linked list?

function deleteNode(head, pos) {
  let current = head;
  for (let i = 0; i < pos - 1; i++) {
    current = current.next;
  }
  current.next = current.next.next;
  return head;
}

medium

A. It does not handle deletion when pos is 0 (head node).

B. It incorrectly updates the head pointer.

C. It causes a syntax error due to missing semicolons.

D. It deletes the last node instead of the specified position.

Solution

1. Step 1: Analyze loop for position traversal

   The loop stops at node before position to delete, which is correct for pos > 0.

2. Step 2: Check handling of pos = 0

   When pos is 0, the loop does not run and current is head, but code does not update head pointer, so deletion fails.

3. Final Answer:

   It does not handle deletion when pos is 0 (head node). -> Option A

4. Quick Check:

   Missing head deletion case causes error [OK]

Hint: Always check if deleting head node separately [OK]

Common Mistakes:

• Ignoring edge case when pos is 0
• Assuming loop handles all positions
• Not updating head pointer when needed

5. You have a singly linked list: 5 -> 15 -> 25 -> 35 -> 45 -> null. You want to delete the node at position 4. What will the list look like after deletion? Also, what happens if you try to delete at position 5?

hard

A. After deletion at position 4: 5 -> 15 -> 25 -> 35 -> null; Deleting position 5 causes a runtime error.

B. After deletion at position 4: 5 -> 15 -> 25 -> 35 -> 45 -> null; Deleting position 5 removes the last node.

C. After deletion at position 4: 5 -> 15 -> 25 -> 35 -> null; Deleting position 5 does nothing as it's out of range.

D. After deletion at position 4: 15 -> 25 -> 35 -> 45 -> null; Deleting position 5 removes the head.

Solution

1. Step 1: Identify node at position 4

   Positions start at 0, so position 4 is the fifth node with value 45.

2. Step 2: Delete node at position 4

   Remove node 45 by linking node 35's next to null.

3. Step 3: Consider deletion at position 5

   Position 5 is out of list range (list length is 5), so deletion does nothing safely.

4. Final Answer:

   After deletion at position 4: 5 -> 15 -> 25 -> 35 -> null; Deleting position 5 does nothing as it's out of range. -> Option C

5. Quick Check:

   Deleting last node removes it; out-of-range deletion ignored [OK]

Hint: Deleting out-of-range position does nothing [OK]

Common Mistakes:

• Trying to delete non-existent node causing errors
• Confusing positions and node values
• Not updating previous node's next pointer correctly