Delete Node at Specific Position in DSA Go - Build from Scratch

Delete Node at Specific Position

📖 Scenario: Imagine you have a list of tasks to do today, and you want to remove a task at a certain position from your list.

🎯 Goal: You will create a singly linked list of tasks, then write code to delete a node at a specific position, and finally print the updated list.

📋 What You'll Learn

Create a singly linked list with exact tasks

Add a variable for the position to delete

Write a function to delete the node at the given position

Print the linked list after deletion

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality, and managing playlists where items need to be added or removed dynamically.

💼 Career

Understanding linked list operations such as deletion at a position is fundamental for software developers working with data structures, improving problem-solving and coding skills.

1

Create the linked list

Create a singly linked list with nodes containing these exact tasks in order: "Task1", "Task2", "Task3", "Task4", "Task5". Use the struct Node with fields data string and next *Node. Create a variable head pointing to the first node.

DSA Go

// Define Node struct and create linked list with tasks
// Your code here

Hint

Start by defining the Node struct. Then create nodes linked by the next pointer.

2

Set the position to delete

Create an integer variable called pos and set it to 2. This means you want to delete the node at position 2 (0-based index) from the linked list.

DSA Go

package main

import "fmt"

type Node struct {
    data string
    next *Node
}

func main() {
    head := &Node{data: "Task1"}
    head.next = &Node{data: "Task2"}
    head.next.next = &Node{data: "Task3"}
    head.next.next.next = &Node{data: "Task4"}
    head.next.next.next.next = &Node{data: "Task5"}

    // Create variable pos and set to 2
    // Your code here
}

Hint

Just create a variable pos and assign it the value 2.

3

Delete the node at position

Write a function called deleteAtPosition that takes head *Node and pos int as parameters and returns the new head of the linked list after deleting the node at position pos. Use 0-based indexing. Then call this function with head and pos and assign the result back to head.

DSA Go

package main

import "fmt"

type Node struct {
    data string
    next *Node
}

func deleteAtPosition(head *Node, pos int) *Node {
    // Your code here
}

func main() {
    head := &Node{data: "Task1"}
    head.next = &Node{data: "Task2"}
    head.next.next = &Node{data: "Task3"}
    head.next.next.next = &Node{data: "Task4"}
    head.next.next.next.next = &Node{data: "Task5"}

    pos := 2

    // Call deleteAtPosition and update head
    // Your code here
}

Hint

Check if pos is 0 to remove the first node. Otherwise, move to the node before pos and change its next pointer to skip the node at pos.

4

Print the updated linked list

Write a for loop to traverse the linked list starting from head. Print each node's data followed by " -> ". After the loop, print "null" to show the end of the list.

DSA Go

package main

import "fmt"

type Node struct {
    data string
    next *Node
}

func deleteAtPosition(head *Node, pos int) *Node {
    if head == nil {
        return head
    }
    if pos == 0 {
        return head.next
    }
    current := head
    for i := 0; current != nil && i  ")
        current = current.next
    }
    fmt.Println("null")
}

Hint

Use a for loop to go through each node and print its data followed by " -> ". End with printing "null".

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