Remove Nth Node From End of List

'''
Method 1: Find length L, remove (L-n)th node from start
         T.C O(n) but 2 pass, S.C O(1)

Method 2: Recursion (count from end while returning)
         T.C O(n), S.C O(n) due to recursion stack

Method 3: Two pointers (FAST-SLOW)  <-- used here
    Move fast pointer n steps ahead.
    If fast becomes None => removing head (n == size), return head.next
    Else move slow and fast together until fast reaches last node.
    Now slow.next is the node to delete.
    T.C O(n) one pass, S.C O(1)
'''
# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
        fast=head
        while fast and n>0:
            fast=fast.next
            n-=1
        if not fast: #remove nth elmenet in a n size linked list
            return head.next
        slow=head
        while fast.next!=None:
            slow=slow.next
            fast=fast.next
        slow.next=slow.next.next
        return head