Insert at Beginning Head Insert in DSA Go - Build from Scratch

Insert at Beginning Head Insert

📖 Scenario: You are managing a simple list of tasks to do. You want to add new tasks at the start of the list so that the newest task is always first.

🎯 Goal: Build a singly linked list in Go and add a new node at the beginning (head insert). Then print the list to see the order of tasks.

📋 What You'll Learn

Define a Node struct with data and next fields

Create a linked list with initial nodes

Write a function to insert a new node at the beginning

Print the linked list nodes in order

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality, and managing dynamic data where insertion order matters.

💼 Career

Understanding linked lists and pointer manipulation is fundamental for software developers working with low-level data structures, memory management, and performance optimization.

1

Create initial linked list nodes

Define a Node struct with data as string and next as *Node. Then create three nodes with data "Task1", "Task2", and "Task3" linked in that order. Set head to point to the first node.

DSA Go

type Node struct {
    data string
    next *Node
}

// Create nodes and link them
// Your code here

Hint

Remember to link nodes by setting the next field to the next node's address.

2

Create a new task to insert

Create a new node called newNode with data "UrgentTask" and next set to nil.

DSA Go

package main

import "fmt"

type Node struct {
    data string
    next *Node
}

func main() {
    node3 := &Node{data: "Task3", next: nil}
    node2 := &Node{data: "Task2", next: node3}
    node1 := &Node{data: "Task1", next: node2}
    head := node1

    // Create newNode with data "UrgentTask"
    // Your code here
}

Hint

Use &Node{data: "UrgentTask", next: nil} to create the new node.

3

Insert the new node at the beginning

Set newNode.next to head. Then update head to point to newNode. This inserts the new node at the start of the list.

DSA Go

package main

import "fmt"

type Node struct {
    data string
    next *Node
}

func main() {
    node3 := &Node{data: "Task3", next: nil}
    node2 := &Node{data: "Task2", next: node3}
    node1 := &Node{data: "Task1", next: node2}
    head := node1

    newNode := &Node{data: "UrgentTask", next: nil}

    // Insert newNode at beginning
    // Your code here
}

Hint

Link the new node to the current head, then update head to newNode.

4

Print the linked list

Use a for loop with a variable current starting at head. Print current.data followed by " -> " until current is nil. After the loop, print "nil".

DSA Go

package main

import "fmt"

type Node struct {
    data string
    next *Node
}

func main() {
    node3 := &Node{data: "Task3", next: nil}
    node2 := &Node{data: "Task2", next: node3}
    node1 := &Node{data: "Task1", next: node2}
    head := node1

    newNode := &Node{data: "UrgentTask", next: nil}

    newNode.next = head
    head = newNode

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

Hint

Use a loop to move through nodes and print their data followed by an arrow.

Practice

1. What does the "Insert at Beginning" operation do in a singly linked list?

easy

A. Adds a new node at the end of the list

B. Searches for a node at the beginning

C. Deletes the first node of the list

D. Adds a new node at the start of the list, making it the new head

Solution

1. Step 1: Understand the head pointer role

   The head pointer points to the first node in the list.

2. Step 2: Insert at beginning changes head

   Inserting at beginning means creating a new node and making it the new head, linking it to the old head.

3. Final Answer:

   Adds a new node at the start of the list, making it the new head -> Option D

4. Quick Check:

   Insert at beginning = new head node [OK]

Hint: New node becomes head, old head moves next [OK]

Common Mistakes:

• Confusing insert at beginning with insert at end
• Thinking it deletes nodes
• Assuming it searches nodes

2. Which of the following Go code snippets correctly inserts a new node with value 10 at the beginning of a singly linked list?

easy

A. newNode := &Node{data: 10, next: head}
head = newNode

B. head.next = &Node{data: 10}
head = head.next

C. newNode := &Node{data: 10}
head.next = newNode

D. head = &Node{data: 10}
head.next = nil

Solution

1. Step 1: Create new node pointing to current head

   newNode is created with data 10 and next pointing to current head.

2. Step 2: Update head to new node

   Assign head to newNode so it becomes the first node.

3. Final Answer:

   newNode := &Node{data: 10, next: head}\nhead = newNode -> Option A

4. Quick Check:

   New node points to old head, head updated [OK]

Hint: New node's next = old head; head = new node [OK]

Common Mistakes:

• Not linking new node to old head
• Changing head.next instead of head
• Forgetting to update head pointer

3. What will be the output linked list after executing this Go code snippet?

package main
import "fmt"

type Node struct {
  data int
  next *Node
}

func main() {
  head := &Node{data: 5, next: nil}
  newNode := &Node{data: 3, next: head}
  head = newNode
  newNode2 := &Node{data: 1, next: head}
  head = newNode2

  // Print list
  current := head
  for current != nil {
    fmt.Print(current.data, " -> ")
    current = current.next
  }
  fmt.Print("null")
}

medium

A. 1 -> 3 -> 5 -> null

B. 5 -> 3 -> 1 -> null

C. 3 -> 1 -> 5 -> null

D. null

Solution

1. Step 1: Insert 3 at beginning

   New node with 3 points to 5, head updated to 3.

2. Step 2: Insert 1 at beginning

   New node with 1 points to 3, head updated to 1.

3. Step 3: Print list from head

   Traverse from head: 1 -> 3 -> 5 -> null.

4. Final Answer:

   1 -> 3 -> 5 -> null -> Option A

5. Quick Check:

   Head insert order reversed list [OK]

Hint: New nodes become head, so list order reverses insert order [OK]

Common Mistakes:

• Printing in insertion order instead of head order
• Confusing next pointers
• Not updating head after insert

4. Identify the error in this Go code that tries to insert a new node with value 7 at the beginning of a linked list:

type Node struct {
  data int
  next *Node
}

func insertAtBeginning(head *Node, val int) *Node {
  newNode := &Node{data: val}
  newNode.next = head.next
  head = newNode
  return head
}

medium

A. The function should not return head

B. head should not be reassigned inside the function

C. newNode.next should point to head, not head.next

D. newNode is not initialized correctly

Solution

1. Step 1: Check newNode.next assignment

   It incorrectly points to head.next instead of head, skipping the current head node.

2. Step 2: Correct pointer to head

   newNode.next must point to head to insert before it.

3. Final Answer:

   newNode.next should point to head, not head.next -> Option C

4. Quick Check:

   Link new node to head, not head.next [OK]

Hint: New node's next = head, not head.next [OK]

Common Mistakes:

• Linking new node to head.next skipping head
• Not updating head pointer correctly
• Ignoring return value of updated head

5. Given a singly linked list with nodes containing integer values, write a Go function to insert multiple values at the beginning in order, so the final list reflects the insertion order. Which approach correctly achieves this?

hard

A. Insert each new value at the end one by one.

B. Insert each new value at the beginning but reverse the input slice before inserting.

C. Insert each new value at the beginning one by one.

D. Insert each new value at the end but reverse the input slice before inserting.

Solution

1. Step 1: Understand insertion order at beginning

   Inserting at beginning reverses order because new nodes become head.

2. Step 2: Reverse input slice before inserting

   Reversing input before inserting at beginning restores original order in final list.

3. Final Answer:

   Insert each new value at the beginning but reverse the input slice before inserting. -> Option B

4. Quick Check:

   Reverse input + head insert = original order [OK]

Hint: Reverse input before head inserts to keep order [OK]

Common Mistakes:

• Inserting at beginning without reversing input
• Inserting at end without reversing input
• Confusing insert at beginning with insert at end