Pop Using Linked List Node in DSA Go - Execution Trace

Concept Flow - Pop Using Linked List Node

Start with head node

↓

Check if head is None?

Yes→List is empty, cannot pop

No↓

Save head node data

↓

Move head pointer to next node

↓

Return saved data

↓

Done

This flow shows how popping removes the first node from the linked list by moving the head pointer to the next node and returning the removed node's data.

Execution Sample

DSA Go

func Pop(head **Node) (int, bool) {
    if *head == nil {
        return 0, false
    }
    popped := (*head).data
    *head = (*head).next
    return popped, true
}

This function removes the first node from the linked list and returns its data if the list is not empty.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with list: 10 -> 20 -> 30 -> nil	10 -> 20 -> 30 -> nil	head points to 10	head -> [10] -> [20] -> [30] -> nil
2	Check if head is nil	10 -> 20 -> 30 -> nil	No change	head -> [10] -> [20] -> [30] -> nil
3	Save data from head node	10 -> 20 -> 30 -> nil	No change	head -> [10] -> [20] -> [30] -> nil
4	Move head pointer to next node	20 -> 30 -> nil	head moves from 10 to 20	head -> [20] -> [30] -> nil
5	Return popped data = 10	20 -> 30 -> nil	No change	head -> [20] -> [30] -> nil
6	End of pop operation	20 -> 30 -> nil	head points to 20	head -> [20] -> [30] -> nil

💡 Pop ends after moving head to next node and returning the removed node's data.

Variable Tracker

Variable	Start	After Step 3	After Step 4	Final
head	points to node with data 10	points to node with data 10	points to node with data 20	points to node with data 20
popped	undefined	10	10	10
list nodes	10 -> 20 -> 30 -> nil	10 -> 20 -> 30 -> nil	20 -> 30 -> nil	20 -> 30 -> nil

Key Moments - 3 Insights

Why do we check if head is nil before popping?

Because if head is nil, the list is empty and there is no node to pop. This is shown in execution_table step 2 where the check prevents errors.

What happens to the removed node after popping?

The removed node is no longer referenced by head, so it can be cleaned up by memory management. The list now starts from the next node, as seen in step 4.

Why do we move head to head.next instead of deleting the node directly?

Moving head to head.next effectively removes the first node from the list. The original node is no longer accessible from the list, which is the goal of pop. This is clear in step 4.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'popped' after step 3?

A 10

B 20

C nil

D 0

💡 Hint

Check the 'popped' variable in variable_tracker after step 3.

At which step does the head pointer move to the next node?

A Step 3

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Pointer Changes' column in execution_table.

If the list was empty initially, what would happen at step 2?

A Head moves to next node

B Popped value is set to 10

C Function returns false and 0

D List remains unchanged

💡 Hint

Refer to the concept_flow and execution_table step 2 for empty list check.

Concept Snapshot

Pop Using Linked List Node:
- Check if list is empty (head == nil)
- Save data from head node
- Move head pointer to next node
- Return saved data
- List shrinks by one node at front

Full Transcript

This concept shows how to remove the first node from a linked list by moving the head pointer to the next node and returning the removed node's data. The process starts by checking if the list is empty. If not, it saves the data from the head node, moves the head pointer forward, and returns the saved data. This effectively removes the first node from the list.

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