# Container With Most Water

## Problem

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.

![](https://1063826111-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MGdx41c9p2PMgIHbUTK%2F-MNPHr-naIuux1wRQMJ0%2F-MNPX37rFv6eVfkJCABH%2Fimage.png?alt=media\&token=ffd3c42c-5cd6-47a8-818c-2b4002282543)

{% hint style="info" %}
For example:

```
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.
```

{% endhint %}

## Solution

```
class Solution:
    def maxArea(self, height: List[int]) -> int:
        maxArea = float("-inf")
        i = 0
        j = len(height) - 1
        
        while i < j:
            thisArea = min(height[i], height[j]) * (j-i)
            maxArea = max(maxArea, thisArea)
            if height[i] < height[j]:
                i+=1
            else:
                j-=1
        return maxArea
        
#Two pointer approach: The two pointer approach can 
#be used as we can set both pointers at the end of the
#array and move either pointer forward depending on
#whether ai or aj is the minimum

#Two pointer is the best approach since we need to
#calculate the width between two points 
```


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