You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example 2:

Input: height = [1,1]
Output: 1

Constraints:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

We will implement this question using opposite two-pointers with the following steps.

1. Initialize left and right pointer to 0 and len(height)-1. Initialize max_area = 0.
2. Perform two pointers selections (steps 3-4) while left < right.
3. The max_area is the maximum of the current max_area and (right - left) * min(height[left], height[right])
4. Update left += 1 if height[left] < height[right], else right -= 1.
5. Return the final max_area.

The logic behind this two pointers algorithm is that we start from the two ends and work our way into the middle. Given two vertical lines, the area of that container is (right - left) * min(height[left], height[right]). So the only way we can hold more water when the horizontal distance between two pointers decrease, is to increase the vertical distance min(height[left], height[right]). We wish to keep the taller line from height[left], height[right] and continue searching for another line that's taller than the lower line, which potentially forms a larger container.

Implementation

def maxArea(self, height: List[int]) -> int:
    left, right = 0, len(height) - 1
    max_area = 0
    while left < right:
        max_area = max(max_area, (right - left) * min(height[left], height[right]))
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_area