Delete Node at Specific Position in DSA Python - Build from Scratch

Delete Node at Specific Position

📖 Scenario: Imagine you have a chain of friends holding hands in a line. Sometimes, one friend needs to leave the line from a specific spot. We want to help by removing that friend without breaking the chain.

🎯 Goal: You will build a simple linked list and write code to remove a friend (node) from a specific position in the chain.

📋 What You'll Learn

Create a linked list with exact nodes: 10, 20, 30, 40, 50

Create a variable called position to hold the position of the node to delete

Write code to delete the node at the given position in the linked list

Print the linked list after deletion in the format: 10 -> 20 -> 40 -> 50 -> None

💡 Why This Matters

🌍 Real World

Linked lists are used in many software systems to manage ordered data where items can be added or removed efficiently, like music playlists or undo history in apps.

💼 Career

Understanding linked lists and how to manipulate them is a fundamental skill for software developers, especially when working with low-level data structures or optimizing performance.

1

Create the linked list

Create a linked list with nodes containing these exact values in order: 10, 20, 30, 40, 50. Use a class called Node with attributes data and next. Create a variable called head that points to the first node.

DSA Python

# Define Node class and create linked list with nodes 10, 20, 30, 40, 50
# Your code here

Hint

Start by defining the Node class with data and next. Then create each node and link them by setting the next attribute.

2

Set the position to delete

Create a variable called position and set it to 3. This means you want to delete the node at position 3 in the linked list (counting starts at 1).

DSA Python

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

head = Node(10)
second = Node(20)
third = Node(30)
fourth = Node(40)
fifth = Node(50)

head.next = second
second.next = third
third.next = fourth
fourth.next = fifth

# Create variable position and set it to 3
# Your code here

Hint

Just create a variable named position and assign the value 3 to it.

3

Delete the node at the given position

Write code to delete the node at the position stored in position from the linked list starting at head. If position is 1, update head to the next node. Otherwise, find the node before the position and change its next to skip the node at position.

DSA Python

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

head = Node(10)
second = Node(20)
third = Node(30)
fourth = Node(40)
fifth = Node(50)

head.next = second
second.next = third
third.next = fourth
fourth.next = fifth

position = 3

# Delete the node at position
# Your code here

Hint

Check if position is 1 to update head. Otherwise, move to the node before the one to delete and change its next to skip the deleted node.

4

Print the linked list after deletion

Write code to print the linked list starting from head in the format: 10 -> 20 -> 40 -> 50 -> None. Use a variable called current to traverse the list and print each node's data followed by ->. End with None.

DSA Python

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

head = Node(10)
second = Node(20)
third = Node(30)
fourth = Node(40)
fifth = Node(50)

head.next = second
second.next = third
third.next = fourth
fourth.next = fifth

position = 3

if position == 1:
    head = head.next
else:
    current = head
    count = 1
    while current is not None and count < position - 1:
        current = current.next
        count += 1
    if current is not None and current.next is not None:
        current.next = current.next.next

# Print the linked list
# Your code here

Hint

Use a loop to go through each node starting at head. Print the data and an arrow until you reach the end, then print None.