[Leetcode] Container With Most Water

[Problem Link]

• Use a typical 2 pointer approach to solve this problem. At step 1 assume your max container spans two ends.
• What is the area of a container given i and j? It is the width of the container (j-i) x height of the container (which is bounded by the min(heights[i], heights[j]).
• Moving which of these pointers might possibly increase the area of our container in the next step? It seems like the minimum of the two heights is holding back so lets move the pointer pointing to min(heights[i], heights[j]).

class Solution {
    public int maxArea(int[] height) {
        int i = 0 ; 
        int j = height.length-1;
        int max = Integer.MIN_VALUE;
 
        while ( i < j ) {
            int currwidth = j-i;
            int currheight = Math.min(height[i], height[j]);
            max = Math.max(max, currwidth*currheight);
 
            if ( height[i] < height[j] )
                i++;
            else
                j--;
        }
 
        return max;
    }
}