## Description

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

Given *n* non-negative integers *a _{1}*,

*a*, …,

_{2}*a*, where each represents a point at coordinate (

_{n }*i*,

*a*).

_{i}*n*vertical lines are drawn such that the two endpoints of line

*i*is at (

*i*,

*a*) and (

_{i}*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.

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

## Explanation

First, draw an x, y-axis picture to convert text information.

Then generalize ideas from the picture.

The container is a rectangle. So the amount of the water is determined by the width and height. The container’s width is right – left and height is the minimum value between left line height and right line height.

## Java Solution

```
class Solution {
public int maxArea(int[] height) {
int maxArea = 0;
if (height == null) {
return maxArea;
}
int left = 0;
int right = height.length - 1;
while (left < right) {
int area = (right - left) * Math.min(height[left], height[right]);
maxArea = Math.max(maxArea, area);
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return maxArea;
}
}
```

## Python Solution

```
class Solution:
def maxArea(self, height: List[int]) -> int:
max_area = 0
i = 0
j = len(height) - 1
while i < j:
area_width = j - i
area_height = min(height[i], height[j])
area = area_width * area_height
max_area = max(max_area, area)
if height[i] < height[j]:
i += 1
else:
j -= 1
return max_area
```

- Time Complexity: ~N
- Space Complexity: ~1

It is a best solution found that very popular and helpful:

https://www.youtube.com/watch?v=P1VeKzXfY-k&list=PL1MJrDFRFiKZi8qmnPEbuF9-2sJmtl5Am&index=130&ab_channel=EricProgramming