Insert at Beginning Head Insert in DSA Go

Mental Model

Add a new item right at the start so it becomes the first thing in the list.

Analogy: Imagine putting a new book on the front of a line of books on a shelf, pushing the others back.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, insert value 0 at beginning

Goal: Add 0 at the start so the list becomes 0 -> 1 -> 2 -> 3 -> null

Step 1: Create new node with value 0

newNode(0) -> null
head -> 1 -> 2 -> 3 -> null

Why: We need a new node to insert at the beginning

Step 2: Set new node's next to current head (node 1)

newNode(0) -> 1 -> 2 -> 3 -> null
head -> 1 -> 2 -> 3 -> null

Why: Link new node to the old first node to keep the list connected

Step 3: Update head to new node

head -> 0 -> 1 -> 2 -> 3 -> null

Why: New node is now the first node in the list

Result:

head -> 0 -> 1 -> 2 -> 3 -> null

Annotated Code

DSA Go

package main

import "fmt"

type Node struct {
	value int
	next  *Node
}

type LinkedList struct {
	head *Node
}

func (ll *LinkedList) InsertAtBeginning(val int) {
	newNode := &Node{value: val}
	newNode.next = ll.head // link new node to old head
	ll.head = newNode      // update head to new node
}

func (ll *LinkedList) PrintList() {
	curr := ll.head
	for curr != nil {
		fmt.Printf("%d -> ", curr.value)
		curr = curr.next
	}
	fmt.Println("null")
}

func main() {
	ll := &LinkedList{}
	// create initial list 1 -> 2 -> 3 -> null
	ll.InsertAtBeginning(3)
	ll.InsertAtBeginning(2)
	ll.InsertAtBeginning(1)

	// insert 0 at beginning
	ll.InsertAtBeginning(0)

	ll.PrintList()
}

newNode := &Node{value: val}

create new node with given value

newNode.next = ll.head // link new node to old head

link new node to current first node

ll.head = newNode // update head to new node

make new node the new head of the list

Output

0 -> 1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(1) because we only change a few pointers without traversing the list

Space: O(1) because we create only one new node regardless of list size

vs Alternative: Compared to inserting at the end which requires O(n) traversal, head insert is faster and simpler

Edge Cases

empty list

new node becomes the head and points to null

DSA Go

ll.head = newNode      // update head to new node

single element list

new node points to the existing node and becomes new head

DSA Go

newNode.next = ll.head // link new node to old head

When to Use This Pattern

When you need to add items quickly at the front of a list, use head insert to avoid traversing the list.

Common Mistakes

Mistake: Forgetting to update the head pointer to the new node
Fix: Always assign the new node to head after linking it to the old head

Mistake: Not linking the new node's next to the old head, breaking the list
Fix: Set newNode.next = head before updating head

Summary

Adds a new node at the start of a linked list.

Use when you want to quickly insert elements at the front without traversal.

The new node points to the old head, then becomes the new head.

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