---
title: '19. Remove Nth Node From End of List'
description: Given the head of a linked list, remove the n^th node from the end of the list and return its head
icon: dot
topics:
  - { name: "Linked List", slug: "linked-list" }
  - { name: "Two Pointers", slug: "two-pointers" }
issue: "https://github.com/prdlk/leetcode/issues/80"
sidebar:
  label: 'Remove Nth Node From End of List'
  badge: 'Medium'
---

::::warning
Could you do this in one pass?
::::

### 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`

## Solution

```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: 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
```