Why Delete Node at Specific Position in DSA Go?

The Big Idea

What if you could remove any item from a list instantly without breaking the chain?

The Scenario

Imagine you have a long paper chain of names, and you want to remove the 5th name. You have to count each name one by one and carefully tear out the right one without breaking the chain.

The Problem

Doing this by hand is slow and easy to mess up. You might tear the wrong link or lose track of the count. It's hard to keep the chain connected properly after removing a name.

The Solution

Using a linked list and deleting a node at a specific position lets a computer quickly find and remove the right link without breaking the chain. It keeps everything connected smoothly and saves time.

Before vs After

✗ Before

names := []string{"Anna", "Bob", "Cara", "Dan", "Eli"}
// To delete 5th name, shift all after it manually
for i := 4; i < len(names)-1; i++ {
    names[i] = names[i+1]
}
names = names[:len(names)-1]

✓ After

type Node struct {
    data string
    next *Node
}

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

What It Enables

This lets programs efficiently remove any item from a list without breaking the order or wasting time.

Real Life Example

Think of a music playlist where you want to remove the 3rd song without stopping the whole list or reordering everything manually.

Key Takeaways

Manual removal is slow and error-prone.

Deleting a node in a linked list keeps connections intact.

It makes removing items fast and safe in many applications.

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