> For the complete documentation index, see [llms.txt](https://jaywin.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jaywin.gitbook.io/leetcode/solutions/0011-container-with-most-water.md).

# 0011. Container With Most Water

<https://leetcode.com/problems/container-with-most-water>

## Description

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 the line `i` is at `(i, ai)` and `(i, 0)`. Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

**Notice** that you may not slant the container.

**Example 1:**

![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)

```
**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
```

**Example 3:**

```
**Input:** height = [4,3,2,1,4]
**Output:** 16
```

**Example 4:**

```
**Input:** height = [1,2,1]
**Output:** 2
```

**Constraints:**

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

## ac1: 2 pointer

start from left & right, move smaller one forward

[https://leetcode.com/problems/container-with-most-water/discuss/6099/Yet-another-way-to-see-what-happens-in-the-O(n)-algorithm](https://leetcode.com/problems/container-with-most-water/discuss/6099/Yet-another-way-to-see-what-happens-in-the-O%28n%29-algorithm)

```java
class Solution {
    public int maxArea(int[] height) {
        // edge cases
        if (height.length < 2) return 0;

        int l = 0, r = height.length - 1, max = 0;
        while (l < r) {
            int vol = (r - l) * Math.min(height[l], height[r]);
            max = Math.max(max, vol); // if current volume > sum, update

            if (height[l] < height[r]) {
                l++;
            } else {
                r--;
            }
        }

        return max;
    }
}
```