Pop Using Linked List Node in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'pop' operation do in a linked list?

It removes the first node (head) from the linked list and returns its value, effectively shortening the list by one node.

beginner

In a singly linked list, what pointer changes when you pop a node?

The head pointer is updated to point to the next node after the current head, removing the first node from the list.

intermediate

Why is popping from a linked list efficient?

Because it only requires updating the head pointer and does not need to traverse the list, making it an O(1) operation.

beginner

What should you check before popping a node from a linked list?

You should check if the list is empty (head is nil) to avoid errors when trying to pop from an empty list.

intermediate

Show the Go code snippet to pop a node from a singly linked list.

func (list *LinkedList) Pop() (int, bool) { if list.Head == nil { return 0, false } value := list.Head.Value list.Head = list.Head.Next return value, true }

What happens to the linked list head after a pop operation?

A It becomes nil always

B It points to the last node

C It points to the next node

D It points to the previous node

After popping, the head moves to the next node in the list.

What is the time complexity of popping a node from the head of a singly linked list?

A O(1)

B O(n)

C O(log n)

D O(n^2)

Popping from the head is a constant time operation because it only updates the head pointer.

What should your pop function return if the linked list is empty?

A An error or false indicating failure

B The last node's value

C Zero always

D The head pointer

You should indicate failure or return a special value when popping from an empty list.

Which pointer is updated during a pop operation in a singly linked list?

A Tail pointer

B Previous pointer

C Next pointer of last node

D Head pointer

The head pointer is updated to the next node after popping.

If the linked list has only one node, what happens after pop?

A Head points to itself

B Head becomes nil

C Tail pointer changes

D List becomes circular

After popping the only node, the list becomes empty, so head is nil.

Explain how the pop operation works on a singly linked list node.

Think about what happens to the head pointer and the node it points to.

Describe the edge cases to consider when popping from a linked list.

What happens if there is no node or only one node?

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