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"""

Remove Nth Node From End of List

Given the head of a linked list, remove the nth node from the end of the list and return its head.

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

Follow up: Could you do this in one pass?

"""

# 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, n):
        fast = head
        slow = head
        # advance fast to nth position
        for i in range(n):
            fast = fast.next

        if not fast:
            return head.next
        # then advance both fast and slow now they are nth postions apart
        # when fast gets to None, slow will be just before the item to be deleted
        while fast.next:
            slow = slow.next
            fast = fast.next
        # delete the node
        slow.next = slow.next.next
        return head

"""

Remove Nth Node From End of List

Given the head of a linked list, remove the nth node from the end of the list and return its head.

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

Follow up: Could you do this in one pass?

"""

# 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, n):
        fast = head
        slow = head
        # advance fast to nth position
        for i in range(n):
            fast = fast.next

        if not fast:
            return head.next
        # then advance both fast and slow now they are nth postions apart
        # when fast gets to None, slow will be just before the item to be deleted
        while fast.next:
            slow = slow.next
            fast = fast.next
        # delete the node
        slow.next = slow.next.next
        return head
Remove Nth Node From End of List  Given the head of a linked list, remove the nth node from the end of the list and return its head.  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  Follow up: Could you do this in one pass? # Definition for singly-linked list. # advance fast to nth position # then advance both fast and slow now they are nth postions apart # when fast gets to None, slow will be just before the item to be deleted # delete the node