LinkedList solution.

removeN.py

# 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]:
        if not head.next:
            # Single node input meaning n = 1
            return None
        
        print(f"Nth node: {n}")
        col = []
        idx = 0
        node = head
        
        while node:
            if len(col) == n + 1:
                col = col[1:]
            
            col += [node]
            node = node.next
        
        s = len(col)
        if n > 1 and n < s:
            col[0].next = col[2] # Normal case
        elif n == 1:
            col[-2].next = None # Drop last
        elif n == s:
            head = col[1] # Drop first
            
        return head