"""
19. Remove Nth Node From End of List
Difficulty: Medium
https://leetcode.com/problems/remove-nth-node-from-end-of-list/

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Given the head of a linked list, remove the n^th node from the end of
the list and return its head.



Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]



Constraints:

        • The number of nodes in the list is sz.

        • 1 <= sz <= 30

        • 0 <= Node.val <= 100

        • 1 <= n <= sz



Follow up: Could you do this in one pass?
"""


# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode | None, n: int) -> ListNode | None:
        dummy = ListNode(0)
        dummy.next = head

        fast = dummy
        for _ in range(n):
            if fast.next is None:
                return head
            fast = fast.next

        slow = dummy
        while fast.next is not None:
            slow = slow.next
            fast = fast.next

        slow.next = slow.next.next

        return dummy.next