# LeetCode 11. Container With Most Water

## Description

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

Given n non-negative integers a1a2, …, a, where each represents a point at coordinate (iai). n vertical lines are drawn such that the two endpoints of line i is at (iai) and (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