Why Delete Node at Beginning in DSA Go?

The Big Idea

What if you could remove the first item instantly without moving everyone else?

The Scenario

Imagine you have a line of people waiting to buy tickets. If the first person leaves, you have to ask everyone else to move forward manually.

The Problem

Manually moving everyone forward is slow and confusing. You might lose track of who is next or accidentally skip someone.

The Solution

Using a linked list, you can simply change the start pointer to the next person, instantly removing the first without moving others.

Before vs After

✗ Before

for i := 1; i < len(queue); i++ {
    queue[i-1] = queue[i]
}
queue = queue[:len(queue)-1]

✓ After

head = head.next

What It Enables

This lets you quickly remove the first item without touching the rest, making operations fast and error-free.

Real Life Example

In a printer queue, when a document finishes printing, it is removed instantly from the front so the next document can start.

Key Takeaways

Manual shifting is slow and error-prone.

Changing the start pointer removes the first node instantly.

This makes data handling efficient and simple.

Practice

1. What happens when you delete the first node in a singly linked list?

easy

A. The head pointer moves to the second node

B. The last node is deleted

C. The list becomes empty regardless of nodes

D. The middle node is deleted

Solution

1. Step 1: Understand the head pointer role

   The head pointer points to the first node in the list.

2. Step 2: Effect of deleting the first node

   Deleting the first node means moving the head pointer to the next node, which becomes the new first node.

3. Final Answer:

   The head pointer moves to the second node -> Option A

4. Quick Check:

   Deleting first node updates head = next node [OK]

Hint: Deleting first node means head points to next node [OK]

Common Mistakes:

• Thinking last node is deleted
• Assuming list becomes empty always
• Deleting middle node by mistake

2. Which of the following Go code snippets correctly deletes the first node of a singly linked list with head pointer head?

easy

A. head.next = nil

B. head.next = head

C. head = nil.next

D. head = head.next

Solution

1. Step 1: Understand pointer update to delete first node

   To delete the first node, head must point to the second node, which is head.next.

2. Step 2: Check each option for correctness

   head = head.next correctly assigns head = head.next. Others either assign incorrectly or cause errors.

3. Final Answer:

   head = head.next -> Option D

4. Quick Check:

   head = head.next deletes first node [OK]

Hint: Assign head to head.next to delete first node [OK]

Common Mistakes:

• Assigning head.next to head (wrong direction)
• Using nil.next which causes error
• Setting head.next to nil deletes wrong node

3. Given the following Go code for a singly linked list, what will be printed after deleting the first node?

package main
import "fmt"

type Node struct {
    value int
    next  *Node
}

func main() {
    head := &Node{value: 1}
    head.next = &Node{value: 2}
    head.next.next = &Node{value: 3}

    // Delete first node
    head = head.next

    // Print list
    current := head
    for current != nil {
        fmt.Print(current.value, " -> ")
        current = current.next
    }
    fmt.Print("null")
}

medium

A. 2 -> 3 -> null

B. 1 -> 2 -> 3 -> null

C. 3 -> null

D. null

Solution

1. Step 1: Initial list nodes and values

   The list starts as 1 -> 2 -> 3 -> null.

2. Step 2: Deleting first node by head = head.next

   Head moves from node with value 1 to node with value 2, so list becomes 2 -> 3 -> null.

3. Final Answer:

   2 -> 3 -> null -> Option A

4. Quick Check:

   After deletion, list starts at second node [OK]

Hint: After deletion, list starts from second node [OK]

Common Mistakes:

• Printing original list without deletion
• Assuming first node still prints
• Confusing null with empty list

4. The following Go code attempts to delete the first node of a singly linked list but causes a runtime error. What is the error and how to fix it?

package main

type Node struct {
    value int
    next  *Node
}

func main() {
    var head *Node = nil

    // Delete first node
    head = head.next
}

medium

A. Compile error: cannot assign head.next to head

B. Runtime panic: nil pointer dereference; fix by checking if head is nil before deletion

C. No error; code works fine

D. Syntax error: missing import; fix by adding fmt package

Solution

1. Step 1: Identify cause of runtime panic

   head is nil, so accessing head.next causes nil pointer dereference panic.

2. Step 2: Fix by checking if head is nil before deleting

   Check if head != nil before assigning head = head.next to avoid panic.

3. Final Answer:

   Runtime panic: nil pointer dereference; fix by checking if head is nil before deletion -> Option B

4. Quick Check:

   Nil check prevents runtime panic [OK]

Hint: Always check head != nil before deleting first node [OK]

Common Mistakes:

• Ignoring nil check causing panic
• Assuming head is never nil
• Confusing syntax and runtime errors

5. You have a singly linked list with nodes: 10 -> 20 -> 30 -> 40 -> null. Write a Go function to delete the first node and return the new head. What will be the list after deletion?

func deleteFirstNode(head *Node) *Node {
    if head == nil {
        return nil
    }
    head = head.next
    return head
}

hard

A. 30 -> 40 -> null

B. 10 -> 20 -> 30 -> 40 -> null

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

D. null

Solution

1. Step 1: Understand function logic

   If head is nil, return nil. Otherwise, move head to head.next and return it.

2. Step 2: Apply function to list 10 -> 20 -> 30 -> 40 -> null

   Deleting first node (10) moves head to node with value 20, so list becomes 20 -> 30 -> 40 -> null.

3. Final Answer:

   20 -> 30 -> 40 -> null -> Option C

4. Quick Check:

   Function returns list starting from second node [OK]

Hint: Return head.next to delete first node and update list [OK]

Common Mistakes:

• Returning original head without deletion
• Not handling nil head case
• Deleting wrong node by mistake