# 19. Remove Nth Node From End of List
## Problem
{% embed url="" %}
## Intuition
## Time Complexity
## Space Complexity
## Solution
```java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
// Time Complexity: O(n), n -> number of nodes in linked list
// Space complexity: O(1), Auxillary space complexity
// Adding sentinel node. Adding this helps in moving the
// nodes using 0-index correctly
// Create a dummy node
ListNode dummy = new ListNode();
dummy.next = head; // Link head to dummy.next
// Create two pointers setting to dummy
ListNode slow = dummy;
ListNode fast = dummy;
// Move fast pointer by "n" steps forward
for(int i=0;i<=n;i++) {
fast = fast.next;
}
// Traverse fast and slow pointers till end of fast ptr
while(fast!=null) {
slow = slow.next;
fast = fast.next;
}
// Rearranging the nodes
slow.next = slow.next.next;
// As head is set to dummy.next, return dummy.next
// Where the head ListNode is manipulated
return dummy.next;
}
}
```
