Creating a Singly Linked List from Scratch in DSA Go - Code Step by Step

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the Node struct with a value and a pointer to the next node.

DSA Go

type Node struct {
    value int
    next [1]Node
}

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

A*

Bpointer

C&

Dref

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pointer' or 'ref' which are not valid Go syntax for pointers.

Using '&' which is the address operator, not a type declaration.

In Go, to declare a pointer to a struct, we use the * symbol before the type name.

2fill in blank

medium

Complete the code to create a new node with a given value.

DSA Go

func NewNode(val int) *Node {
    return &Node{value: [1]
}

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Avalue

Bval

Cval.value

DNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' which is the field name, not the parameter.

Using 'val.value' which is invalid since val is an int.

The function parameter is named val, so we assign it to the value field of the Node struct.

3fill in blank

hard

Fix the error in the Append method to add a new node at the end of the list.

DSA Go

func (head *Node) Append(val int) {
    current := head
    for current.next != [1] {
        current = current.next
    }
    current.next = NewNode(val)
}

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Ahead

Bcurrent

C0

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 0 or head instead of nil.

Using current instead of current.next in the loop condition.

The loop should continue until current.next is nil, meaning the end of the list.

4fill in blank

hard

Fill both blanks to complete the PrintList method that prints all node values.

DSA Go

func (head *Node) PrintList() {
    current := head
    for current != [1] {
        fmt.Print(current.[2], " -> ")
        current = current.next
    }
    fmt.Println("nil")
}

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Anil

Bvalue

Cval

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'val' instead of 'value' for the field name.

Checking for 'head' instead of 'nil' in the loop condition.

The loop runs until current is nil, and the field to print is 'value'.

5fill in blank

hard

Fill all three blanks to complete the Remove method that deletes the first node with a given value.

DSA Go

func (head *Node) Remove(val int) *Node {
    if head == [1] {
        return head
    }
    if head.value == val {
        return head.[2]
    }
    current := head
    for current.next != [3] {
        if current.next.value == val {
            current.next = current.next.next
            break
        }
        current = current.next
    }
    return head
}

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

Anil

Bnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'head' instead of 'nil' for empty list check.

Returning head instead of head.next when removing the first node.

Looping until current.next equals head instead of nil.

Check if head is nil, return head; if head.value matches val, return head.next; loop until current.next is nil.

Practice

1. What is the main purpose of a singly linked list in data structures?

easy

A. To store elements in a hash table for fast lookup

B. To store elements in a tree-like structure with multiple branches

C. To store elements in a fixed-size array for quick access

D. To store elements in a linear order where each element points to the next

Solution

1. Step 1: Understand singly linked list structure

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike trees or arrays, singly linked lists have a linear sequence with pointers only to the next element.

3. Final Answer:

   To store elements in a linear order where each element points to the next -> Option D

4. Quick Check:

   Singly linked list = linear chain with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing singly linked list with trees or arrays
• Thinking nodes have pointers to previous nodes
• Assuming fixed size storage like arrays

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

easy

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

B. type Node struct { value int; prev *Node }

C. type Node struct { value int; next Node }

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

Solution

1. Step 1: Identify correct pointer usage

   In Go, the next node must be a pointer to Node, so *Node is correct.

2. Step 2: Check struct fields

   Field 'value' stores data, 'next' points to the next node. 'prev' is for doubly linked lists, and 'interface{}' is too generic.

3. Final Answer:

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

4. Quick Check:

   Node next pointer = *Node [OK]

Hint: Use *Node for next pointer in Go struct [OK]

Common Mistakes:

• Using value type instead of pointer for next
• Confusing singly linked list with doubly linked list (prev pointer)
• Using interface{} instead of *Node for next

3. What will be the output of the following Go code that creates a singly linked list with nodes 1 -> 2 -> 3?

package main
import "fmt"
type Node struct {
  value int
  next  *Node
}
func main() {
  head := &Node{value: 1}
  second := &Node{value: 2}
  third := &Node{value: 3}
  head.next = second
  second.next = third
  current := head
  for current != nil {
    fmt.Print(current.value)
    if current.next != nil {
      fmt.Print(" -> ")
    }
    current = current.next
  }
  fmt.Println(" -> null")
}

medium

A. 1 -> 2 -> null

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

C. 1 -> 2 -> 3 -> null

D. Compilation error

Solution

1. Step 1: Trace node creation and linking

   Nodes with values 1, 2, 3 are created and linked in order: head.next = second, second.next = third.

2. Step 2: Trace printing loop

   Loop prints current.value, then ' -> ' if next exists, moving from head to third, then prints ' -> null'.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Print nodes in order 1,2,3 with arrows [OK]

Hint: Follow next pointers from head to end [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing last node in output
• Confusing pointer assignments causing nil errors

4. Identify the error in this Go code snippet that tries to append a new node with value 4 to the end of a singly linked list:

func appendNode(head *Node, val int) {
  newNode := &Node{value: val}
  current := head
  for current.next != nil {
    current = current.next
  }
  current = newNode
}

medium

A. The loop condition should be current != nil instead of current.next != nil

B. The new node is not linked to the list; current pointer is reassigned incorrectly

C. The new node should be created with next pointing to head

D. The function should return the new head pointer

Solution

1. Step 1: Analyze the loop and pointer update

   The loop correctly finds the last node where current.next is nil.

2. Step 2: Check how new node is linked

   Assigning current = newNode only changes the local variable, not the last node's next pointer. It should be current.next = newNode.

3. Final Answer:

   The new node is not linked to the list; current pointer is reassigned incorrectly -> Option B

4. Quick Check:

   Link new node by current.next = newNode [OK]

Hint: Assign new node to current.next, not current [OK]

Common Mistakes:

• Reassigning current instead of current.next
• Wrong loop condition causing nil pointer dereference
• Not returning updated head when needed

5. You want to create a function in Go that reverses a singly linked list. Which approach correctly reverses the list by changing the next pointers?

func reverseList(head *Node) *Node {
  var prev *Node = nil
  current := head
  for current != nil {
    nextTemp := current.next
    current.next = prev
    prev = current
    current = nextTemp
  }
  return prev
}

What will be the output if the original list is 1 -> 2 -> 3 -> null and you print the reversed list?

hard

A. 3 -> 2 -> 1 -> null

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

C. null

D. Runtime error due to nil pointer

Solution

1. Step 1: Understand reversal logic

   The function iterates through the list, reversing each node's next pointer to point to the previous node.

2. Step 2: Trace reversal on list 1 -> 2 -> 3

   After reversal, the list becomes 3 -> 2 -> 1 -> null, and prev points to new head.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list head = prev = 3 [OK]

Hint: Reverse pointers by tracking prev, current, next [OK]

Common Mistakes:

• Not updating current.next correctly
• Losing track of next node causing infinite loop
• Returning old head instead of new head