Remove Nth Node from End of List in DSA Python - Build from Scratch

Remove Nth Node from End of List

📖 Scenario: You are managing a queue of customers waiting in line at a store. Sometimes, a customer leaves the line from the end, not just the front. You want to remove the nth customer from the end of the line.

🎯 Goal: Build a program that removes the nth node from the end of a singly linked list and prints the updated list.

📋 What You'll Learn

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

Create a variable n to specify which node from the end to remove

Implement the logic to remove the nth node from the end of the list

Print the linked list after removal in the format: 10 -> 20 -> 30 -> 50 -> null

💡 Why This Matters

🌍 Real World

Removing the nth node from the end of a list is useful in managing queues, undo histories, or any ordered data where you need to remove elements relative to the end.

💼 Career

This technique is commonly asked in coding interviews and is important for understanding linked list manipulation and pointer techniques.

1

Create the linked list

Create a singly linked list with nodes containing these exact values in order: 10, 20, 30, 40, 50. Use the class ListNode provided below and create a variable called head pointing to the first node.

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

# Create linked list nodes and connect them
# Your code here

Hint

Start by creating the first node with value 10 and assign it to head. Then link each next node with the given values.

2

Set the position to remove

Create a variable called n and set it to 2. This means you want to remove the 2nd node from the end of the list.

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

head = ListNode(10)
head.next = ListNode(20)
head.next.next = ListNode(30)
head.next.next.next = ListNode(40)
head.next.next.next.next = ListNode(50)

# Set n to 2
# Your code here

Hint

Just create a variable n and assign it the value 2.

3

Remove the nth node from the end

Write code to remove the nth node from the end of the linked list starting at head. Use two pointers called fast and slow. Move fast ahead by n nodes first, then move both pointers until fast reaches the end. Remove the node after slow. Handle the case when the node to remove is the head.

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

head = ListNode(10)
head.next = ListNode(20)
head.next.next = ListNode(30)
head.next.next.next = ListNode(40)
head.next.next.next.next = ListNode(50)

n = 2

# Remove the nth node from the end
# Your code here

Hint

Use a dummy node before head to handle edge cases. Move fast pointer n steps ahead, then move both fast and slow until fast reaches the last node. Then remove the node after slow.

4

Print the updated linked list

Print the linked list starting from head in the format: 10 -> 20 -> 30 -> 50 -> null. Use a loop to traverse the list and print each node's value followed by ->. End with null.

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

head = ListNode(10)
head.next = ListNode(20)
head.next.next = ListNode(30)
head.next.next.next = ListNode(40)
head.next.next.next.next = ListNode(50)

n = 2

dummy = ListNode(0, head)
fast = dummy
slow = dummy

for _ in range(n):
    fast = fast.next

while fast.next:
    fast = fast.next
    slow = slow.next

slow.next = slow.next.next

head = dummy.next

# Print the linked list
# Your code here

Hint

Use a while loop to go through each node starting from head. Print the value and -> until the end. Then print null.