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

Insert at Beginning (Head Insert) in a Linked List

📖 Scenario: Imagine you are managing a line of people waiting to enter a concert. Sometimes, a VIP arrives and needs to be placed right at the front of the line. We will simulate this using a linked list where each person is a node.

🎯 Goal: You will build a simple linked list and learn how to insert a new node at the beginning (head) of the list. This means the new person will be first in line.

📋 What You'll Learn

Create a Node class with data and next attributes

Create a LinkedList class with a head attribute

Write a method insert_at_beginning to add a new node at the start

Print the linked list after insertion to show the order

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like music playlists, undo functionality in apps, and managing queues where order matters.

💼 Career

Understanding linked lists and how to insert nodes is fundamental for software developers, especially when working with dynamic data structures and memory management.

1

Create Node and LinkedList classes

Create a class called Node with an __init__ method that takes data and sets self.data = data and self.next = None. Then create a class called LinkedList with an __init__ method that sets self.head = None.

DSA Python

# Define Node class
# Define LinkedList class
# Your code here

Hint

Think of Node as a person in line holding data and a pointer to the next person. The LinkedList keeps track of the first person with head.

2

Add insert_at_beginning method

Inside the LinkedList class, define a method called insert_at_beginning that takes self and data. Create a new Node with data. Set the new node's next to the current head. Then update head to the new node.

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    # Define insert_at_beginning method here
    # Your code here

Hint

Remember, the new node points to the old first node, then becomes the new first node.

3

Add method to print the linked list

Inside the LinkedList class, define a method called print_list that takes self. Use a variable current starting at head. Use a while loop to print current.data followed by ' -> ' and move current to current.next. After the loop, print None.

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_at_beginning(self, data):
        new_node = Node(data)
        new_node.next = self.head
        self.head = new_node

    # Define print_list method here
    # Your code here

Hint

Traverse the list from head to end, printing each node's data followed by an arrow.

4

Insert nodes and print the list

Create a LinkedList object called ll. Use ll.insert_at_beginning(30), then ll.insert_at_beginning(20), then ll.insert_at_beginning(10). Finally, call ll.print_list() to display the list.

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_at_beginning(self, data):
        new_node = Node(data)
        new_node.next = self.head
        self.head = new_node

    def print_list(self):
        current = self.head
        while current:
            print(str(current.data) + ' -> ', end='')
            current = current.next
        print('None')

# Create linked list and insert nodes
# Your code here

Hint

Insert 30 first, then 20 before it, then 10 before 20. The list should print from 10 to 30.

Practice

1. What does the operation insert at beginning do in a linked list?

easy

A. Deletes the first node of the list

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

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

D. Deletes the last node of the list

Solution

1. Step 1: Understand the meaning of 'insert at beginning'

   Inserting at the beginning means adding a new element before the current first element.

2. Step 2: Identify the effect on the linked list

   The new node becomes the new head, pushing the old head to second position.

3. Final Answer:

   Adds a new node at the start of the list -> Option B

4. Quick Check:

   Insert at beginning = Add new head [OK]

Hint: Insert at beginning means new head node [OK]

Common Mistakes:

• Confusing insert at beginning with insert at end
• Thinking it deletes nodes instead of adding
• Assuming it changes the last node

2. Which of the following is the correct syntax to insert a new node at the beginning of a singly linked list in Python?

easy

A. new_node = head.next; head = new_node

B. head.next = new_node; new_node = head

C. head = head.next; new_node = head

D. new_node.next = head; head = new_node

Solution

1. Step 1: Understand pointer assignment for head insert

   To insert at beginning, new_node's next points to current head.

2. Step 2: Update head to new_node

   Then head is updated to new_node to make it the first node.

3. Final Answer:

   new_node.next = head; head = new_node -> Option D

4. Quick Check:

   Set new_node.next to head, then head = new_node [OK]

Hint: Link new node to head, then update head [OK]

Common Mistakes:

• Assigning head.next instead of new_node.next
• Not updating head to new_node
• Swapping assignment order incorrectly

3. What will be the linked list after executing the following code?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(10)
new_node = Node(20)
new_node.next = head
head = new_node

medium

A. 20 -> 10 -> null

B. 10 -> 20 -> null

C. 20 -> null

D. 10 -> null

Solution

1. Step 1: Create initial list with head node 10

   Initially, head points to node with data 10 and next is None.

2. Step 2: Insert new_node with data 20 at beginning

   new_node.next points to head (node 10), then head updated to new_node.

3. Final Answer:

   20 -> 10 -> null -> Option A

4. Quick Check:

   New node 20 becomes head, points to 10 [OK]

Hint: New node points to old head, then becomes head [OK]

Common Mistakes:

• Thinking new_node is added at end
• Not updating head to new_node
• Confusing node data order

4. Identify the error in this code snippet that tries to insert a node at the beginning of a linked list:

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(5)
new_node = Node(15)
head = new_node
new_node.next = head

medium

A. The order of assignments is wrong; new_node.next should be set before head update

B. Node class is missing a data attribute

C. head should be None initially

D. new_node should not be created before head

Solution

1. Step 1: Analyze assignment order

   head is set to new_node first, then new_node.next is set to head (which is now new_node itself).

2. Step 2: Identify logical error

   This causes new_node.next to point to itself, losing original list starting at old head.

3. Final Answer:

   The order of assignments is wrong; new_node.next should be set before head update -> Option A

4. Quick Check:

   Set new_node.next before head = new_node [OK]

Hint: Set new_node.next before updating head [OK]

Common Mistakes:

• Updating head before linking new_node.next
• Confusing attribute initialization
• Assuming head must be None initially

5. Given a linked list with nodes containing values [30, 20, 10] (30 is head), which sequence of operations correctly inserts a new node with value 40 at the beginning and results in the list 40 -> 30 -> 20 -> 10 -> null?

hard

A. new_node = Node(40); new_node.next = None; head = new_node

B. new_node = Node(40); head.next = new_node; head = new_node

C. new_node = Node(40); new_node.next = head; head = new_node

D. new_node = Node(40); head = new_node; new_node.next = head

Solution

1. Step 1: Create new node with value 40

   new_node is created with data 40 and next initially None.

2. Step 2: Link new_node to current head

   Set new_node.next to current head (node 30) to keep existing list after new node.

3. Step 3: Update head to new_node

   Head pointer is updated to new_node, making it the first node.

4. Final Answer:

   new_node = Node(40); new_node.next = head; head = new_node -> Option C

5. Quick Check:

   Link new node to old head, then update head [OK]

Hint: Link new node to head, then update head pointer [OK]

Common Mistakes:

• Linking head.next instead of new_node.next
• Updating head before linking new_node.next
• Setting new_node.next to None losing list