Delete Node at Specific Position in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to delete the head node when position is 0.

DSA Javascript

if (position === 0) {
  head = head[1];
  return head;
}

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

A.child

B.next

C.prev

D.down

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev instead of .next

Trying to delete without updating head

Using incorrect property names

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

2fill in blank

medium

Complete the code to traverse the list to the node before the one to delete.

DSA Javascript

let current = head;
for (let i = 0; i < position - 1; i++) {
  current = current[1];
}

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

A.down

B.prev

C.child

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev which goes backward

Using incorrect property names

Not looping enough times

To move forward in the list, we use the .next property to go to the next node.

3fill in blank

hard

Fix the error in the code to correctly delete the node at the given position.

DSA Javascript

if (current && current.next) {
  current.next = current.next[1];
}

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

A.prev

B.down

C.next

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev which is backward link

Using wrong property names

Not checking if current.next exists

To delete the node, we link the current node's next to the node after the one being deleted using .next.

4fill in blank

hard

Fill both blanks to complete the function that deletes a node at a specific position.

DSA Javascript

function deleteNode(head, position) {
  if (position === 0) {
    head = head[1];
    return head;
  }
  let current = head;
  for (let i = 0; i < position - 1; i++) {
    current = current[2];
  }
  if (current && current.next) {
    current.next = current.next.next;
  }
  return head;
}

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

A.next

B.prev

C.child

D.down

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev or other properties instead of .next

Not updating head correctly when deleting the first node

To delete the node, move head to head.next if position is 0, and traverse using .next to reach the node before the one to delete.

5fill in blank

hard

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

DSA Javascript

function deleteNode(head, position) {
  if (position === 0) {
    head = head[1];
    return head;
  }
  let current = head;
  for (let i = 0; i < position - 1; i++) {
    current = current[2];
  }
  if (current && current.next) {
    current.next = current.next[3];
  }
  return head;
}

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

A.next

B.prev

C.next.next

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev or .child instead of .next

Not linking current.next to current.next.next correctly

To delete the node, move head to head.next if position is 0, traverse using .next, and link current.next to current.next.next to skip the deleted node.

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