Pop Using Linked List Node in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to remove the first node from the linked list.

DSA Go

func (list *LinkedList) Pop() *Node {
    if list.Head == nil {
        return nil
    }
    poppedNode := list.Head
    list.Head = list.Head.[1]
    return poppedNode
}

Drag options to blanks, or click blank then click option'

AChild

BNext

CPrev

DParent

Attempts:

3 left

💡 Hint

Common Mistakes

Using Prev instead of Next causes the list to move backward.

Using Child or Parent is incorrect as they are not linked list pointers.

The Pop method removes the first node by moving the head pointer to the next node. So, list.Head.Next is assigned to list.Head.

2fill in blank

medium

Complete the code to check if the linked list is empty before popping.

DSA Go

func (list *LinkedList) Pop() *Node {
    if list.Head == [1] {
        return nil
    }
    poppedNode := list.Head
    list.Head = list.Head.Next
    return poppedNode
}

Drag options to blanks, or click blank then click option'

Anil

B0

Cfalse

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if list.Head == 0 is incorrect because head is a pointer, not a number.

Using true or false is incorrect as head is not a boolean.

The linked list is empty if the head pointer is nil, so we check if list.Head == nil.

3fill in blank

hard

Fix the error in the Pop method to update the tail pointer when popping the last node.

DSA Go

func (list *LinkedList) Pop() *Node {
    if list.Head == nil {
        return nil
    }
    poppedNode := list.Head
    list.Head = list.Head.Next
    if list.Head == [1] {
        list.Tail = [2]
    }
    return poppedNode
}

Drag options to blanks, or click blank then click option'

Anil

Blist.Head

Clist.Tail

DpoppedNode

Attempts:

3 left

💡 Hint

Common Mistakes

Setting tail to poppedNode keeps a reference to removed node.

Setting tail to list.Head when head is nil causes errors.

After popping, if the head becomes nil, it means the list is empty, so tail should also be set to nil.

4fill in blank

hard

Fill both blanks to correctly update the linked list after popping the first node.

DSA Go

func (list *LinkedList) Pop() *Node {
    if list.Head == nil {
        return nil
    }
    poppedNode := list.Head
    list.Head = list.Head.[1]
    if list.Head == nil {
        list.Tail = [2]
    }
    return poppedNode
}

Drag options to blanks, or click blank then click option'

ANext

BPrev

CpoppedNode

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Using Prev instead of Next moves backward incorrectly.

Setting tail to poppedNode keeps removed node reference.

We move head to head.Next and if head is nil, tail must be set to nil to indicate empty list.

5fill in blank

hard

Fill all three blanks to safely pop the first node and update pointers.

DSA Go

func (list *LinkedList) Pop() *Node {
    if list.Head == [1] {
        return nil
    }
    poppedNode := list.Head
    list.Head = list.Head.[2]
    if list.Head == [3] {
        list.Tail = nil
    }
    return poppedNode
}

Drag options to blanks, or click blank then click option'

Anil

BNext

CpoppedNode

Dlist.Head

Attempts:

3 left

💡 Hint

Common Mistakes

Not checking for nil head causes runtime errors.

Using poppedNode instead of nil when updating tail.

Using list.Head instead of nil in condition.

Check if head is nil before popping, move head to head.Next, and if head is nil after popping, set tail to nil.

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