# 11. Container With Most Water

You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.

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: ```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:

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

Solution
we will do this using two pointer approach. we will make a start pointer and an end pointer. The start will be at 0 and the end will be at the last position. Now we will find the area in that container that will be
area=distance between them x minimum of heights of start and end
Now we will move the pointer which is smaller so that we can get a better answer than this.
We will do this until start<end.
Here is the code -

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