Good afternoon! Here's our prompt for today.

Removing the Nth Node From a Linked List's End: A Single-Pass Approach

Get ready to sail through a classic problem: Removing the nth node from the end of a linked list in a single pass. By the end of this lesson, you'll be able to wave goodbye to that pesky nth node and still maintain the list's structure.

The Task

Imagine a chain of dominoes, each representing a node in our linked list. In this scenario, a list might look like this:

1 -> 2 -> 3 -> 4 -> 5

Your mission is to remove the nth domino from the end. So, if we call removeFromEnd(head, 2), the chain would lose the 4 domino, and the head would still be 1.

Ground Rules

Before you jump in, let's set some ground rules:

• Empty List Handling: The linked list could be empty, akin to a chain with no dominoes.
• Nth Node Edge Case: If n exceeds the length of the list, then the chain remains intact. No dominoes are toppled!
• Efficiency Requirements: Aim for a time complexity of O(n) and a space complexity of O(1).

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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def remove_from_end(head, idx_from_end):

    # fill in

    return

​

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# Node definition

class Node:

    def __init__(self, val):

        self.val = val

        self.next = None

​

​

def create_nodes(head, nodes):

    for val in nodes:

        new_node = Node(val)

        head.next = new_node

        head = new_node

​

​

list1 = Node(3)

nodes1 = [4, 5, 6, 7, 8, 9, 10]

create_nodes(list1, nodes1)

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list2 = Node(1)

nodes2 = [2, 3, 4, 5, 6, 7, 8]

create_nodes(list2, nodes2)

​

​

def list_to_str(head):

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We'll now take you through what you need to know.

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