Container with most water

We will take two pointers, one at the beginning and another at the end. Then we calculate the water in between by taking the min value of this 2 position and multiply by the difference of these pointers. We will keep a running max result. Then we move the pointer from the lowest number and move it by 1. Then repeat the process again until 2 pointers meet.

class Solution:
    def maxArea(self, height: List[int]) -> int:
        left, right = 0 , len(height)-1
        res = 0

        while left < right:
            res = max(res, min(height[left], height[right]) * (right - left))
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1
        return res

We traverse the height list only once, so time complexity is O(n). We only use one variable to keep track of our max result, so space complexity is O(1).