<p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], n = 2
<strong>Output:</strong> [1,2,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [1], n = 1
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2], n = 1
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is <code>sz</code>.</li>
	<li><code>1 &lt;= sz &lt;= 30</code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= sz</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do this in one pass?</p>

# 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]:
        newList = ListNode()
        newList.next = head
        fast_pointer = newList
        slow_pointer = newList
        n=n+1                                       # need to be on the n-1th node to skip nth node

        if n == 0:
            return head

        while n and fast_pointer:                   # advance fast_pointer n steps ahead
            fast_pointer = fast_pointer.next
            n-=1
        
        while fast_pointer:                         # advance both to the end
            fast_pointer = fast_pointer.next
            slow_pointer = slow_pointer.next
        
        slow_pointer.next = slow_pointer.next.next

        return head if slow_pointer != newList else slow_pointer.next
        

# nl 0 1 2 3 4 5 6  n = 2
# sf

16000