Delete Node at Specific Position in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting node at position 2

What is the printed linked list after deleting the node at position 2 (0-based index)?

DSA Go

package main
import "fmt"
type Node struct {
    data int
    next *Node
}
func deleteAtPosition(head *Node, pos int) *Node {
    if head == nil {
        return nil
    }
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; current != nil && i < pos-1; i++ {
        current = current.next
    }
    if current == nil || current.next == nil {
        return head
    }
    current.next = current.next.next
    return head
}
func printList(head *Node) {
    for head != nil {
        fmt.Printf("%d -> ", head.data)
        head = head.next
    }
    fmt.Println("null")
}
func main() {
    head := &Node{data: 1}
    head.next = &Node{data: 2}
    head.next.next = &Node{data: 3}
    head.next.next.next = &Node{data: 4}
    head = deleteAtPosition(head, 2)
    printList(head)
}

A 1 -> 2 -> 3 -> null

B 2 -> 3 -> 4 -> null

C 1 -> 2 -> 4 -> null

D 1 -> 3 -> 4 -> null

💡 Hint

Remember that position 2 means the third node in the list.

Deleting the node at position 2 removes the node with data 3. The list becomes 1 -> 2 -> 4 -> null.

❓ Predict Output

intermediate

Output after deleting node at position 0

What is the printed linked list after deleting the node at position 0 (head node)?

DSA Go

package main
import "fmt"
type Node struct {
    data int
    next *Node
}
func deleteAtPosition(head *Node, pos int) *Node {
    if head == nil {
        return nil
    }
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; current != nil && i < pos-1; i++ {
        current = current.next
    }
    if current == nil || current.next == nil {
        return head
    }
    current.next = current.next.next
    return head
}
func printList(head *Node) {
    for head != nil {
        fmt.Printf("%d -> ", head.data)
        head = head.next
    }
    fmt.Println("null")
}
func main() {
    head := &Node{data: 10}
    head.next = &Node{data: 20}
    head.next.next = &Node{data: 30}
    head = deleteAtPosition(head, 0)
    printList(head)
}

A 30 -> null

B 10 -> 20 -> 30 -> null

C null

D 20 -> 30 -> null

💡 Hint

Deleting position 0 removes the head node.

Deleting the head node (position 0) removes the node with data 10. The list becomes 20 -> 30 -> null.

🔧 Debug

advanced

Identify the error in deletion at position function

What error will this code produce when trying to delete a node at position 3 in a list of length 3?

DSA Go

package main
import "fmt"
type Node struct {
    data int
    next *Node
}
func deleteAtPosition(head *Node, pos int) *Node {
    if head == nil {
        return nil
    }
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; i < pos-1; i++ {
        current = current.next
    }
    current.next = current.next.next
    return head
}
func main() {
    head := &Node{data: 5}
    head.next = &Node{data: 10}
    head.next.next = &Node{data: 15}
    head = deleteAtPosition(head, 3)
    fmt.Println("Deletion done")
}

A runtime error: invalid memory address or nil pointer dereference

B compilation error: missing return statement

C no error, deletion successful

D runtime error: index out of range

💡 Hint

Check if the code verifies the current or current.next is nil before accessing.

The code does not check if current or current.next is nil before accessing current.next.next, causing a nil pointer dereference at position 3 in a list of length 3.

❓ Predict Output

advanced

Output after deleting node at invalid position

What is the printed linked list after trying to delete a node at position 5 in a list of length 4?

DSA Go

package main
import "fmt"
type Node struct {
    data int
    next *Node
}
func deleteAtPosition(head *Node, pos int) *Node {
    if head == nil {
        return nil
    }
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; current != nil && i < pos-1; i++ {
        current = current.next
    }
    if current == nil || current.next == nil {
        return head
    }
    current.next = current.next.next
    return head
}
func printList(head *Node) {
    for head != nil {
        fmt.Printf("%d -> ", head.data)
        head = head.next
    }
    fmt.Println("null")
}
func main() {
    head := &Node{data: 7}
    head.next = &Node{data: 14}
    head.next.next = &Node{data: 21}
    head.next.next.next = &Node{data: 28}
    head = deleteAtPosition(head, 5)
    printList(head)
}

A 7 -> 14 -> 21 -> 28 -> null

B 7 -> 14 -> 21 -> null

C 14 -> 21 -> 28 -> null

D null

💡 Hint

Deleting at a position outside the list length should not change the list.

Position 5 is invalid for a list of length 4, so the list remains unchanged: 7 -> 14 -> 21 -> 28 -> null.

🧠 Conceptual

expert

Why is it important to check for nil before deletion?

In the deleteAtPosition function, why must we check if current or current.next is nil before deleting a node?

A To improve the speed of the deletion operation

B To avoid runtime errors caused by accessing memory that does not exist

C To ensure the list is sorted after deletion

D To automatically update the length of the list

💡 Hint

Think about what happens if you try to access a property of a nil pointer.

Accessing current.next or current.next.next when current or current.next is nil causes a runtime error (nil pointer dereference). Checking prevents this error.

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