# 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: Optional[ListNode], n: int) -> Optional[ListNode]: dummy = ListNode(0, head) left = dummy right = head #we actually want right to be head + n #setting right to be head + n while n > 0 and right: right = right.next n -= 1 #this should get left at previous and right at after thanks to dummy node we inserted while right: left = left.next right = right.next #delete node left.next = left.next.next return dummy.next #bottom solution is what I first came up with and works in leetcode except for if head = [1] and n = 1 which is the case I tried to account for in the second elif statement but it gave me an error saying cannot return [] even though it was fine for the first if statement #big O: is still O(n) just like 1st scenario class Solution: def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]: if head is None: return [] elif head.next is None and n == 1: return [] else: current = head counter = 0 #will become length of LL #to determine length of LL while current is not None: counter += 1 current = current.next #consider adding an edge case here if n > counter #second pass now knowing length of LL previous, current = head, head new_counter = 0 target = counter - n while new_counter is not target + 1: if new_counter == (target - 1): previous = current if new_counter == target: after = current.next previous.next = after current.next = None break current = current.next new_counter += 1 return head

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