> For the complete documentation index, see [llms.txt](https://code.taozirui.com/lc/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://code.taozirui.com/lc/data-structure/array/two-pointer/container-with-most-water.md).

# 11. 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.

![](/files/-Lnqh8AgZ19MeFICSIh3)

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:**

```
Input: [1,8,6,2,5,4,8,3,7]

Output: 49
```

**Thoughts:**

1. Two pointer: the shorter height need to move since the rest solution is guaranteed less optimal

```python
class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        left, right = 0, len(height) -1

        res = 0
        while left <= right:
            res = max(res, (right - left) * min(height[left], height[right]))
            if height[right] > height[left]:
                left += 1
            else:
                right -= 1

        return res
```

```python
class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        res, i, j = 0 , 0, len(height) -1

        while i < j:
            h = min(height[i], height[j])
            res = max(res, (j - i)*h)
            while i < j and height[i] <= h: i+= 1
            while i < j and height[j] <= h: j-= 1

        return res
```