Delete Node at Specific Position in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

func deleteHead(head *Node) *Node {
    if head == nil {
        return nil
    }
    return [1]
}

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

Ahead.next

Bhead

Cnil

Dhead.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head instead of head.next

Returning nil which deletes the whole list

To delete the head node, we return the next node as the new head.

2fill in blank

medium

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

DSA Go

func deleteAtPosition(head *Node, pos int) *Node {
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; i < [1]; i++ {
        current = current.next
    }
    // deletion logic follows
    return head
}

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Apos * 2

Bpos

Cpos + 1

Dpos - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Looping pos times causing off-by-one error

Looping pos + 1 times causing out of range

We stop at the node before the one to delete, so we loop pos - 1 times.

3fill in blank

hard

Fix the error in the deletion step to correctly remove the node at position pos.

DSA Go

func deleteAtPosition(head *Node, pos int) *Node {
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; i < pos-1; i++ {
        current = current.next
    }
    current.next = [1]
    return head
}

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Anil

Bcurrent.next.next

Ccurrent.next

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning current.next to current.next (no change)

Assigning nil which truncates the list

To delete the node at pos, link current.next to the node after the one being deleted.

4fill in blank

hard

Fill both blanks to safely delete a node at position pos, checking for nil pointers.

DSA Go

func deleteAtPosition(head *Node, pos int) *Node {
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; i < [1]; i++ {
        if current == nil {
            return head
        }
        current = current.next
    }
    if current == nil || current.next == nil {
        return head
    }
    current.next = [2]
    return head
}

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Apos - 1

Bpos

Ccurrent.next.next

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Looping pos times causing nil pointer panic

Assigning current.next = nil truncates list

Loop pos - 1 times to reach the node before deletion. Then link current.next to current.next.next to delete.

5fill in blank

hard

Fill all three blanks to implement a complete delete function that handles edge cases.

DSA Go

func deleteAtPosition(head *Node, pos int) *Node {
    if head == nil {
        return nil
    }
    if pos == 0 {
        return [1]
    }
    current := head
    for i := 0; i < [2]; i++ {
        if current == nil {
            return head
        }
        current = current.next
    }
    if current == nil || current.next == nil {
        return head
    }
    current.next = [3]
    return head
}

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

Ahead.next

Bpos - 1

Ccurrent.next.next

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Not checking for nil head

Incorrect loop count

Incorrect pointer assignment

Return head.next if deleting first node. Loop pos - 1 times to reach previous node. Link current.next to current.next.next to delete.

Practice

1. What does deleting a node at a specific position in a linked list mean?

easy

A. Adding a new node at that position

B. Removing the node at that position and linking its previous node to the next node

C. Swapping the node at that position with the last node

D. Counting the number of nodes before that position

Solution

1. Step 1: Understand node deletion in linked list

   Deleting a node means removing it and connecting the previous node to the next node to keep the list intact.

2. Step 2: Identify what happens at the specific position

   The node at the given position is removed, and the previous node's next pointer is updated to skip the deleted node.

3. Final Answer:

   Removing the node at that position and linking its previous node to the next node -> Option B

4. Quick Check:

   Delete node = remove and link neighbors [OK]

Hint: Delete means remove node and link neighbors [OK]

Common Mistakes:

• Thinking deletion adds a node
• Swapping nodes instead of deleting
• Only counting nodes without deletion

2. Which of the following Go code snippets correctly deletes a node at position pos in a singly linked list starting with head?

easy

A. curr := head; for i := 0; i <= pos; i++ { curr = curr.next }; curr.next = curr.next.next

B. head = head.next; for i := 0; i < pos; i++ { head = head.next }

C. if pos == 0 { head = head.next } else { curr := head; for i := 0; i < pos-1; i++ { curr = curr.next }; curr.next = curr.next.next }

D. for i := 0; i < pos; i++ { head = head.next }; head = head.next.next

Solution

1. Step 1: Check deletion at position 0

   If position is 0, we update head to head.next to remove the first node.

2. Step 2: Traverse to node before position and update links

   For other positions, we move to node before the target and set its next to skip the deleted node.

3. Final Answer:

   if pos == 0 { head = head.next } else { curr := head; for i := 0; i < pos-1; i++ { curr = curr.next }; curr.next = curr.next.next } -> Option C

4. Quick Check:

   Correct traversal and link update = if pos == 0 { head = head.next } else { curr := head; for i := 0; i < pos-1; i++ { curr = curr.next }; curr.next = curr.next.next } [OK]

Hint: Update previous node's next to skip deleted node [OK]

Common Mistakes:

• Not handling position 0 separately
• Looping one extra time causing nil pointer
• Incorrectly updating head inside loop

3. What is the output linked list after deleting the node at position 2 from the list: 10 -> 20 -> 30 -> 40 -> 50 -> null?

type Node struct {
  data int
  next *Node
}

// Assume head points to 10
// Delete node at position 2
func deleteAtPosition(head *Node, pos int) *Node {
  if pos == 0 {
    return head.next
  }
  curr := head
  for i := 0; i < pos-1; i++ {
    curr = curr.next
  }
  curr.next = curr.next.next
  return head
}

medium

A. 10 -> 20 -> 40 -> 50 -> null

B. 10 -> 20 -> 30 -> 50 -> null

C. 20 -> 30 -> 40 -> 50 -> null

D. 10 -> 30 -> 40 -> 50 -> null

Solution

1. Step 1: Identify node at position 2

   Position 2 node is the third node with value 30.

2. Step 2: Update links to remove node 30

   Node at position 1 (20) next pointer is updated to node at position 3 (40), skipping 30.

3. Final Answer:

   10 -> 20 -> 40 -> 50 -> null -> Option A

4. Quick Check:

   Deleted 30 at pos 2 = 10->20->40->50 [OK]

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

Common Mistakes:

• Deleting wrong node index
• Not updating links correctly
• Returning wrong head pointer

4. The following Go code tries to delete a node at position pos but causes a runtime error. What is the error and how to fix it?

func deleteAtPosition(head *Node, pos int) *Node {
  curr := head
  for i := 0; i < pos; i++ {
    curr = curr.next
  }
  curr.next = curr.next.next
  return head
}

medium

A. Logic error because loop goes to pos instead of pos-1; fix loop to i < pos-1

B. Syntax error due to missing return statement; fix by adding return head

C. No error; code works fine

D. Runtime error due to nil pointer when pos is 0; fix by handling pos 0 separately

Solution

1. Step 1: Analyze loop traversal

   The loop moves curr to the node at position pos, but we need the node before pos to update links.

2. Step 2: Fix loop to stop at node before pos

   Change loop condition to i < pos-1 so curr points to previous node, allowing safe link update.

3. Final Answer:

   Logic error because loop goes to pos instead of pos-1; fix loop to i < pos-1 -> Option A

4. Quick Check:

   Loop to pos-1 for previous node [OK]

Hint: Loop to node before position, not at position [OK]

Common Mistakes:

• Looping to pos causing nil pointer
• Not handling pos 0 case
• Assuming curr.next always exists

5. Given a singly linked list with nodes: 5 -> 10 -> 15 -> 20 -> 25 -> null, write the resulting list after deleting nodes at positions 0, 2, and then 1 in that order. What is the final list?

hard

A. 10 -> 20 -> null

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

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

D. 10 -> 25 -> null

Solution

1. Step 1: Delete node at position 0

   Remove first node (5), list becomes 10 -> 15 -> 20 -> 25 -> null.

2. Step 2: Delete node at position 2

   Positions now: 0=10,1=15,2=20,3=25. Remove node 20, list becomes 10 -> 15 -> 25 -> null.

3. Step 3: Delete node at position 1

   Positions now: 0=10,1=15,2=25. Remove node 15, list becomes 10 -> 25 -> null.

4. Final Answer:

   10 -> 25 -> null -> Option D

5. Quick Check:

   Delete 0,2,1 in order = 10->25 [OK]

Hint: Update list after each deletion before next [OK]

Common Mistakes:

• Not updating positions after deletions
• Deleting wrong nodes due to stale positions
• Confusing zero-based indexing