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

Time Complexity: Delete Node at Specific Position

O(n)

Understanding Time Complexity

When we delete a node at a specific position in a linked list, we want to know how the time it takes changes as the list grows.

We ask: How much work does the computer do when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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: The for-loop that moves through the list to reach the node before the one to delete.
• How many times: It runs up to position - 1 times, depending on the position given.

How Execution Grows With Input

As the position to delete moves further in the list, the loop runs more times, so the work grows with position.

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

Pattern observation: The number of steps grows roughly in a straight line with the position number.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Deleting a node at any position always takes the same short time."

[OK] Correct: Because to delete a node, you must first reach it by moving through the list, which takes longer for nodes farther down.

Interview Connect

Understanding how time grows with list size helps you explain your code clearly and shows you know what affects performance in real tasks.

Self-Check

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

Practice

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

easy

A. The list remains unchanged

B. The last node is removed

C. The head node is removed and the next node becomes the new head

D. All nodes are deleted

Solution

1. Step 1: Understand position 1 in linked list

   Position 1 refers to the head node of the list.

2. Step 2: Delete the head node

   Removing the head means the next node becomes the new head.

3. Final Answer:

   The head node is removed and the next node becomes the new head -> Option C

4. Quick Check:

   Deleting position 1 removes head [OK]

Hint: Deleting position 1 means removing the head node [OK]

Common Mistakes:

• Thinking position 1 means last node
• Assuming list stays same after deletion
• Deleting all nodes instead of one

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

easy

A. if (pos == 1) head = head.next;

B. head = head.next.next;

C. head.next = null;

D. head = null;

Solution

1. Step 1: Check deletion at position 1

   If position is 1, we remove the head by pointing head to head.next.

2. Step 2: Validate syntax correctness

   if (pos == 1) head = head.next; correctly assigns head = head.next; others either skip nodes or nullify incorrectly.

3. Final Answer:

   if (pos == 1) head = head.next; -> Option A

4. Quick Check:

   Correct syntax for deleting head node [OK]

Hint: Deleting head means head = head.next [OK]

Common Mistakes:

• Skipping two nodes instead of one
• Setting head.next to null incorrectly
• Setting entire head to null without condition

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

medium

A. 10 -> 20 -> 40 -> null

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

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

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Identify node at position 3

   Position 3 node is 30 in the list 10 -> 20 -> 30 -> 40 -> null.

2. Step 2: Remove node 30 and link previous to next

   Link node 20's next to node 40, skipping 30.

3. Final Answer:

   10 -> 20 -> 40 -> null -> Option A

4. Quick Check:

   Deleted 3rd node 30 correctly [OK]

Hint: Skip the node at position by linking previous to next [OK]

Common Mistakes:

• Deleting wrong node position
• Not updating previous node's next pointer
• Removing head instead of position 3

4. Consider this code snippet to delete a node at position pos in a singly linked list:

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

What error can occur if pos is 1?

medium

A. ArrayIndexOutOfBoundsException

B. NullPointerException because temp.next is null

C. The head node is deleted correctly

D. No error, deletion works fine

Solution

1. Step 1: Analyze loop for pos = 1

   Loop runs from i=1 to pos-2 = -1, so it does not run; temp = head.

2. Step 2: Access temp.next when pos=1

   temp.next is head.next, but deletion logic expects previous node; deleting head this way causes NullPointerException.

3. Final Answer:

   NullPointerException because temp.next is null -> Option B

4. Quick Check:

   Deleting position 1 needs special handling [OK]

Hint: Deleting head needs separate code; loop fails for pos=1 [OK]

Common Mistakes:

• Assuming loop runs for pos=1
• Not handling head deletion separately
• Expecting no error for pos=1

5. You have a singly linked list: 5 -> 10 -> 15 -> 20 -> 25 -> null. Write the resulting list after deleting the node at position 5, then position 1, and finally position 2 in that order.

hard

A. 10 -> 20 -> 25 -> null

B. 10 -> 15 -> 25 -> null

C. 5 -> 15 -> 20 -> 25 -> null

D. 10 -> 20 -> null

Solution

1. Step 1: Delete node at position 5

   Position 5 node is 25; after deletion list: 5 -> 10 -> 15 -> 20 -> null.

2. Step 2: Delete node at position 1

   Delete head (5); new list: 10 -> 15 -> 20 -> null.

3. Step 3: Delete node at position 2

   Position 2 node is 15; after deletion list: 10 -> 20 -> null.

4. Final Answer:

   10 -> 20 -> null -> Option D

5. Quick Check:

   Sequential deletions update list correctly [OK]

Hint: Delete from end, then head, then middle stepwise [OK]

Common Mistakes:

• Deleting wrong nodes after list changes
• Not updating head after first deletion
• Confusing positions after deletions