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

Recall & Review

beginner

What does 'Delete Node at Specific Position' mean in a linked list?

It means removing the node that is at a given position (like 1st, 2nd, 3rd) from the linked list, and then linking the previous node to the next node to keep the list connected.

beginner

What should you check before deleting a node at a specific position?

You should check if the position is valid (not less than 1 and not greater than the list length) and if the list is not empty.

beginner

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

To delete the first node, move the head pointer to the next node. The old head node is then removed from the list.

intermediate

What happens to the links when you delete a node in the middle of a linked list?

The previous node's next pointer is updated to point to the node after the one being deleted, skipping over the deleted node.

beginner

Why is it important to handle edge cases when deleting a node at a specific position?

Because positions might be invalid or the list might be empty, and failing to handle these can cause errors or crashes.

What should you do if the position to delete is 1 in a singly linked list?

A Do nothing

B Move the head pointer to the next node

C Delete the last node

D Swap the first two nodes

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

If the position to delete is greater than the length of the list, what should happen?

A Delete the first node

B Delete the last node

C Ignore the request or return an error

D Delete a random node

Positions beyond the list length are invalid, so the operation should be ignored or an error returned.

When deleting a node in the middle, what pointer needs to be updated?

A The next pointer of the previous node

B The head pointer

C The next pointer of the deleted node

D The tail pointer

The previous node's next pointer must point to the node after the deleted node.

What is the first step before deleting a node at a specific position?

A Check if the list is empty or position is valid

B Delete the node immediately

C Print the list

D Reverse the list

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

What happens if you try to delete a node from an empty list?

A A new node is added

B The list crashes

C The list is reversed

D Nothing happens or error is returned

Deleting from an empty list should do nothing or return an error safely.

Explain step-by-step how to delete a node at a specific position in a singly linked list.

Think about what happens at the head, middle, and invalid positions.

What are the key edge cases to consider when deleting a node at a specific position?

Consider list size and position validity.

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