# LC 11. Container With Most Water

### [Problem link](https://leetcode.com/problems/container-with-most-water/)

###### tags: `leedcode` `medium` `c++` `Two Pointer`

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

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:**
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg" style="width: 600px; height: 287px;" />

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:**

- <code>n == height.length</code>
- <code>2 <= n <= 10⁵</code>
- <code>0 <= height[i] <= 10⁴</code>

## Solution 1 - Two Pointer
#### C++

class Solution {
public:
    int maxArea(vector<int>& height) {
        int left = 0;
        int right = height.size() - 1;
        int ans = 0;
        while (left < right) {
            int area = (right - left) * min(height[left], height[right]);
            ans = max(ans, area);
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return ans;
    }
}; 

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
[ref](https://www.bilibili.com/video/BV1Qg411q7ia/?spm_id_from=333.788&vd_source=088937f16fb413336c0cb260ed86a1c3)