Container With Most Water

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.

Your program should return an integer which corresponds to the maximum area of water that can be contained ( Yes, we know maximum area instead of maximum volume sounds weird. But this is 2D plane we are working with for simplicity ).

Input : [1, 5, 4, 3]
Output : 6

Explanation : 5 and 3 are distance 2 apart. So size of the base = 2. Height of container = min(5, 3) = 3. 
So total area = 3 * 2 = 6

Method:

This implies that if there was a better solution possible, it will definitely have height greater than min(a1, aN).

B * H > (N-1) * min(a1, aN)
We know that, B < (N-1)
So, H > min(a1, aN)

This means that we can discard min(a1, aN) from our set and look to solve this problem again from the start.
If a1 < aN, then the problem reduces to solving the same thing for a2, aN.
Else, it reduces to solving the same thing for a1, aN-1

Solution:

Time: O(n)
Space: O(1)

public class Solution {
    public int maxArea(ArrayList<Integer> A) {
        int left = 0;
        int right = A.size() - 1;
        int max = 0;
        while (left < right) {
            int leftVal = A.get(left);
            int rightVal = A.get(right);
            max = Math.max(max, Math.min(leftVal, rightVal) * (right - left));
            if (leftVal < rightVal) {
                left ++;
            } else {
                right --;
            }
            if (left < 0 || right >= A.size()) break;
        }
        return max;
    }
}