class Solution:
    def maxArea(self, height: List[int]) -> int:
        left, right = 0, len(height) - 1 
        
        maxArea = 0 
        
        while left < right: 
            heightOne = height[left] 
            heightTwo = height[right] 
            
            area = min(heightOne, heightTwo) * (right - left) 
            maxArea = max(area, maxArea) 
            
            if heightTwo >= heightOne: #if heights are equal, it doesn't matter if you move left or right pointer 
                left += 1 
            elif heightTwo < heightOne: 
                right -= 1 
            
        return maxArea 
        
        
# BRUTE FORCE 
#         maxArea = 0 

#         for i in range(len(height)): 
#             for secondI in range(i + 1, len(height)): 
#                 heightOne = height[i]
#                 heightTwo = height[secondI]
#                 area = min(heightOne, heightTwo) * (secondI - i) 
#                 maxArea = max(area, maxArea) 

#         return maxArea 

# 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