human
I found an interesting problem on https://leetcode.com/problems/container-with-most-water/:
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python code to find two lines, which together with x-axis forms a container, such that the container contains the most water.

class Solution:
     def maxArea(self, height):
         """
         :type height: List[int]
         :rtype: int
         """
         area = (len(height)-1)*min([height[0], height[-1]])
         left_idx = 0
         right_idx = len(height)-1
         while left_idx < right_idx:
             if height[left_idx] < height[right_idx]:
                 start = left_idx + 1
                 while start < right_idx:
                     if height[start] > height[left_idx]:
                         left_idx = start
                         area2 = (right_idx - left_idx) * min([height[left_idx], height[right_idx]])
                         if area2 > area:
                             area = area2
                         break
                     start += 1
                 if start != left_idx:
                     break
             elif height[left_idx] > height[right_idx]:
                 start = right_idx - 1
                 while start > left_idx:
                     if height[start] > height[right_idx]:
                         right_idx = start
                         area2 = (right_idx - left_idx) * min([height[left_idx], height[right_idx]])
                         if area2 > area:
                             area = area2
                         break
                     start -= 1
                 if start!= right_idx:
                     break
             else:
                 left_idx += 1
         
         return area