Remove N-th Node from End of Linked List

function removeNthFromEnd(head, n) {
  let dummy = new ListNode(0);
  dummy.next = head;
  let first = dummy, second = dummy;
  for(let i=0;i<=n;i++) first = first.next;
  while(first){ first = first.next; second = second.next; }
  second.next = second.next.next;
  return dummy.next;
}
                  

Explanation: Use two pointers with a gap of n nodes to remove the N-th node in one pass.

⏱ O(n) | 💾 O(1)

class Solution {
  public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy, second = dummy;
    for(int i=0;i<=n;i++) first = first.next;
    while(first != null){ first = first.next; second = second.next; }
    second.next = second.next.next;
    return dummy.next;
  }
}
                  

Explanation: Two pointers approach ensures single pass removal of N-th node from end.

⏱ O(n) | 💾 O(1)

def remove_nth_from_end(head, n):
    dummy = ListNode(0)
    dummy.next = head
    first = second = dummy
    for _ in range(n+1): first = first.next
    while first:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next
                  

Explanation: Maintain two pointers n nodes apart to delete target node in one pass.

⏱ O(n) | 💾 O(1)