# Container With Most Water

Updated: Mar 25, 2021

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.

**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

**Example 3:**

**Input****:** height = [4,3,2,1,4]
**Output****:** 16

**Example 4:**

**Input****:** height = [1,2,1]
**Output****:** 2

**Constraints:**

n == height.length

2 <= n <= 3 * 104

0 <= height[i] <= 3 * 104

**Solution:**

**Brute Force:**

```
class Solution {
public int maxArea(int[] height) {
int area=1,result=0;
for(int i=0;i<height.length;i++)
{
for(int j=i+1;j<height.length;j++)
{
area = (j-i) * Math.min(height[j],height[i]);
result = Math.max(result,area);
}
}
return result;
}
}
```

**Linear:**

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