Delete Node at Specific Position in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to delete the head node of a singly linked list.

DSA Java

if (head == null) return null;
head = head[1];

Drag options to blanks, or click blank then click option'

Aprev

Bnext

Cchild

Dlast

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes errors because singly linked lists don't have 'prev'.

Trying to assign head to 'child' or 'last' which are not valid references.

To delete the head node, we move the head pointer to the next node.

2fill in blank

medium

Complete the code to traverse the list to the node before the one to delete at position pos.

DSA Java

Node current = head;
for (int i = 1; i < [1] - 1 && current != null; i++) {
    current = current.next;
}

Drag options to blanks, or click blank then click option'

Aindex

Blength

Csize

Dpos

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' or 'size' instead of the position variable causes incorrect traversal.

Looping until 'pos' instead of 'pos - 1' skips the correct node.

We traverse until the node before the position to delete, so we use 'pos'.

3fill in blank

hard

Fix the error in deleting the node at position pos by correctly updating the next pointer.

DSA Java

if (current == null || current.next == null) return head;
current.next = current.next[1];

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cchild

Dlast

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or 'child' causes errors because those references don't exist in singly linked lists.

Not updating the pointer correctly leads to memory leaks or incorrect list structure.

To delete the node at position pos, link current node to the node after the next one using 'next'.

4fill in blank

hard

Fill both blanks to complete the method that deletes a node at position pos in a singly linked list.

DSA Java

public Node deleteAtPosition(Node head, int pos) {
    if (head == null) return null;
    if (pos == 1) {
        head = head[1];
        return head;
    }
    Node current = head;
    for (int i = 1; i < pos - 1 && current != null; i++) {
        current = current[2];
    }
    if (current == null || current.next == null) return head;
    current.next = current.next.next;
    return head;
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or 'child' instead of 'next' causes errors.

Not updating pointers correctly leads to broken list.

To delete the head, move head to head.next. To traverse, move current to current.next.

5fill in blank

hard

Fill all three blanks to complete the method that deletes a node at position pos and returns the updated list.

DSA Java

public Node deleteAtPosition(Node head, int pos) {
    if (head == null) return null;
    if (pos == 1) {
        head = head[1];
        return head;
    }
    Node current = head;
    for (int i = 1; i < pos - 1 && current != null; i++) {
        current = current[2];
    }
    if (current == null || current.next == null) return head;
    current.next = current.next[3];
    return head;
}

Drag options to blanks, or click blank then click option'

Anext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' causes errors because singly linked lists don't have backward links.

Not updating pointers correctly breaks the list.

All blanks require 'next' to move forward or skip nodes in the singly linked list.

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