# 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;

}

};

Thanks for reading.

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