Algorithm Notes
Summary: Remove Nth Node From End — notes not yet curated.
Time: Estimate via loops/recurrences; common classes: O(1), O(log n), O(n), O(n log n), O(n^2)
Space: Count auxiliary structures and recursion depth.
Tip: See the Big-O Guide for how to derive bounds and compare trade-offs.
Big-O Guide
Source
"""
Remove Nth Node From End
TODO: Add problem description
"""
from src.interview_workbook.leetcode._nodes import ListNode
from src.interview_workbook.leetcode._registry import register_problem
from src.interview_workbook.leetcode._types import Category, Difficulty
class Solution:
def solve(self, *args):
"""Removes the nth node from end using two pointers and returns head."""
head, n = args
dummy = ListNode(0, head)
first = second = dummy
# Advance first by n+1 steps
for _ in range(n + 1):
first = first.next
# Move both until first reaches end
while first:
first = first.next
second = second.next
# Remove node
second.next = second.next.next
return dummy.next
def demo():
"""Builds a linked list [1,2,3,4,5], removes 2nd from end, returns list as str."""
# Build linked list
head = ListNode(1)
current = head
for i in range(2, 6):
current.next = ListNode(i)
current = current.next
# Remove nth node
solver = Solution()
new_head = solver.solve(head, 2)
# Convert to list string
result = []
while new_head:
result.append(str(new_head.val))
new_head = new_head.next
return "[" + ",".join(result) + "]"
register_problem(
id=19,
slug="remove-nth-node-from-end-of-list",
title="Remove Nth Node From End of List",
category=Category.TWO_POINTERS,
difficulty=Difficulty.MEDIUM,
tags=["Linked List", "Two Pointers"],
url="https://leetcode.com/problems/remove-nth-node-from-end-of-list/",
)