Pop Using Linked List Node in DSA Go - Build from Scratch

Pop Using Linked List Node

📖 Scenario: Imagine you have a stack of books represented as a linked list. You want to remove the top book from the stack.

🎯 Goal: Build a simple linked list and write code to remove (pop) the first node, then print the remaining list.

📋 What You'll Learn

Create a linked list with three nodes containing values 10, 20, and 30 in that order

Create a variable called head pointing to the first node

Write code to remove the first node (pop) from the linked list

Print the linked list values after popping the first node

💡 Why This Matters

🌍 Real World

Linked lists are used in many programs to manage data where items are added or removed frequently, like undo history or playlist management.

💼 Career

Understanding how to manipulate linked lists is important for software development roles that involve data structure optimization and memory management.

1

Create the linked list nodes

Create three linked list nodes named node1, node2, and node3 with values 10, 20, and 30 respectively. Link them so that node1 points to node2 and node2 points to node3. Create a variable head that points to node1.

DSA Go

package main

type Node struct {
    value int
    next  *Node
}

func main() {
    // Create nodes and link them
    // Your code here
}

Hint

Use &Node{value: X} to create nodes and set next pointers to link them.

2

Prepare to pop the first node

Create a variable called poppedNode and set it to the current head node. This will hold the node we want to remove.

DSA Go

package main

type Node struct {
    value int
    next  *Node
}

func main() {
    node1 := &Node{value: 10}
    node2 := &Node{value: 20}
    node3 := &Node{value: 30}

    node1.next = node2
    node2.next = node3

    head := node1

    // Set poppedNode to head
    // Your code here
}

Hint

Assign head to poppedNode to keep track of the node to remove.

3

Pop the first node from the linked list

Update the head variable to point to the next node after head. This removes the first node from the list.

DSA Go

package main

type Node struct {
    value int
    next  *Node
}

func main() {
    node1 := &Node{value: 10}
    node2 := &Node{value: 20}
    node3 := &Node{value: 30}

    node1.next = node2
    node2.next = node3

    head := node1

    poppedNode := head

    // Update head to next node
    // Your code here
}

Hint

Set head to head.next to remove the first node.

4

Print the linked list after popping

Use a for loop with a variable current starting at head to print all node values separated by -> . End with null.

DSA Go

package main

type Node struct {
    value int
    next  *Node
}

func main() {
    node1 := &Node{value: 10}
    node2 := &Node{value: 20}
    node3 := &Node{value: 30}

    node1.next = node2
    node2.next = node3

    head := node1

    poppedNode := head

    head = head.next

    // Print linked list values
    // Your code here
}

Hint

Use a loop to print each node's value followed by -> , then print null at the end.

Practice

1. What does the pop operation do in a linked list when implemented using nodes?

easy

A. Counts the total number of nodes

B. Adds a new node at the end

C. Removes the first node and returns its value

D. Searches for a node with a specific value

Solution

1. Step 1: Understand pop operation in linked list

   Pop means removing the top or first element from the list.

2. Step 2: Identify what happens to the first node

   The first node is removed and its value is returned, updating the head pointer.

3. Final Answer:

   Removes the first node and returns its value -> Option C

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop always removes the first node in a linked list [OK]

Common Mistakes:

• Thinking pop adds nodes instead of removing
• Confusing pop with search or count operations
• Removing the last node instead of the first

2. Which of the following is the correct Go syntax to define a node struct for a singly linked list?

easy

A. class Node { int value; Node next; }

B. struct Node { int value; Node* next; }

C. Node := struct { value int; next Node }

D. type Node struct { value int; next *Node }

Solution

1. Step 1: Recall Go struct syntax

   In Go, structs are defined with 'type Name struct { fields }'.

2. Step 2: Check pointer usage for next node

   The next field must be a pointer to Node: '*Node'.

3. Final Answer:

   type Node struct { value int; next *Node } -> Option D

4. Quick Check:

   Go struct with pointer = type Node struct { value int; next *Node } [OK]

Hint: Use 'type' and pointers for linked list nodes in Go [OK]

Common Mistakes:

• Using C/C++ or Java syntax in Go
• Missing pointer '*' for next node
• Trying to declare struct inline incorrectly

3. What will be the output after popping once from the linked list represented by nodes with values 10 -> 20 -> 30 -> null?

medium

A. 20 -> 30 -> null

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

C. 30 -> null

D. null

Solution

1. Step 1: Identify the first node value

   The first node has value 10.

2. Step 2: Remove first node and update head

   After popping, head points to node with value 20, so list is 20 -> 30 -> null.

3. Final Answer:

   20 -> 30 -> null -> Option A

4. Quick Check:

   Pop removes 10, new head 20 [OK]

Hint: Pop removes first node, so list starts from second node [OK]

Common Mistakes:

• Not updating head pointer after pop
• Returning the whole list instead of updated list
• Confusing pop with removing last node

4. Given this Go code snippet for pop operation, what is the error?

func pop(head *Node) (*Node, int) {
    if head == nil {
        return nil, 0
    }
    value := head.value
    head = head.next
    return head, value
}

medium

A. It returns wrong value from the node

B. The function does not update the original head pointer outside

C. It causes a runtime panic due to nil dereference

D. The function should return only the value, not the new head

Solution

1. Step 1: Understand pointer behavior in Go

   Assigning 'head = head.next' changes local copy, not original head outside.

2. Step 2: Identify missing pointer update

   The original head pointer outside the function remains unchanged, so pop has no effect.

3. Final Answer:

   The function does not update the original head pointer outside -> Option B

4. Quick Check:

   Local pointer change doesn't affect caller [OK]

Hint: Update original pointer by returning new head or using double pointer [OK]

Common Mistakes:

• Assuming local pointer change updates original
• Ignoring need to return new head
• Not checking for nil before accessing value

5. You have a linked list with nodes 5 -> 15 -> 25 -> 35 -> null. You pop twice and then add a new node with value 45 at the front. What is the final list?

hard

A. 45 -> 25 -> 35 -> null

B. 25 -> 35 -> 45 -> null

C. 45 -> 35 -> null

D. 15 -> 25 -> 35 -> 45 -> null

Solution

1. Step 1: Pop twice from the list

   First pop removes 5, list becomes 15 -> 25 -> 35 -> null. Second pop removes 15, list becomes 25 -> 35 -> null.

2. Step 2: Add new node with value 45 at front

   New node 45 points to current head 25, so list is 45 -> 25 -> 35 -> null.

3. Final Answer:

   45 -> 25 -> 35 -> null -> Option A

4. Quick Check:

   Pop twice then add front = 45 -> 25 -> 35 [OK]

Hint: Pop removes from front; adding front inserts new head [OK]

Common Mistakes:

• Not updating head after each pop
• Adding new node at end instead of front
• Confusing order of operations