<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>

class Solution:
    def removeNthFromEnd(self, h: Optional[ListNode], n: int) -> Optional[ListNode]:
        cnt=0
        temp=h

        while temp:
            cnt+=1
            temp=temp.next
        
        if (cnt==n):return h.next

        temp=h
        for i in range(cnt-n-1):
            temp=temp.next
        
        if temp.next:
            temp.next=temp.next.next
        
        return h


        
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