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 line i is at (i, ai) 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
class Solution {
public int maxArea(int[] height) {
}
}
Idea - 1
We try all of the possible
class Solution {
public int maxArea(int[] height) {
int max = 0;
for(int i = 0; i < height.length-1; ++i)
{
for(int j = i+1; j < height.length; ++j)
{
int curr = Math.min(height[i], height[j])*(j-i);
max = Math.max(curr, max);
}
}
return max;
}
}
Runtime: 206 ms, faster than 20.38% of Java online submissions for Container With Most Water.
Memory Usage: 41.3 MB, less than 5.05% of Java online submissions for Container With Most Water.
Idea - 2
We start at two ends because that maximize width. The minimum of the two bars would define the height of the container - this area is best possible for the shorter container. So we move past the shorter. Time
class Solution {
public int maxArea(int[] height) {
int i = 0, j = height.length-1;
int max = 0;
while(i < j)
{
int w = j-i;
int h = Math.min(height[i], height[j]);
max = Math.max(max, w*h);
if(height[i] == h)
{
++i;
}
else
{
--j;
}
}
return max;
}
}
Runtime: 2 ms, faster than 97.80% of Java online submissions for Container With Most Water.
Memory Usage: 40.6 MB, less than 22.88% of Java online submissions for Container With Most Water.