Container with most water

class Main {
    public static void main(String[] args) {
        /**
         * https://leetcode.com/problems/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.
         * 
         * 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.
         */
         
         int[] height = {1,8,6,2,5,4,8,3,7};
         
         System.out.println(maxArea(height));
    }
    
    public static int maxArea(int[] height) {
        int left = 0;
        int right = height.length - 1;
        
        // maxArea will keep the greatest area calculated and return it at the end
        int maxArea = 0;
        
        while (left < right) {
            // We can capture the index of the lower area
            int lowerHeightIndex = height[left] < height[right] ? left : right;
            
            int lowerHeight = height[lowerHeightIndex];
            // We can calculate the area with the lower element between height[left] and height[right] by 
            // the amount of heights in between the left and the right
            int area = lowerHeight * (right - left);
            if (maxArea < area) {
                maxArea = area;
            }
            
            // We need to move the pointer from the lower height index according to the 
            // pointer referenced. For example: if the lower height came from the left pointer
            // then we need to move the left pointer one position (left++). Otherwise, we need to move the
            // right pointer one position (right--).
            if (lowerHeightIndex == left) {
                left++;
            } else {
                right--;
            }
        }
        
        return maxArea;
    }
}