ContainerWithMostWater.java

/**


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 ).

    Note: You may not slant the container

Example :

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.


**/

public class Solution {
    public int maxArea(int[] A) {
        int pointer1 = 0;
        int pointer2 = A.length - 1;
        int maximum = 0;
        if(pointer1 == pointer2)
        {
            return 0;
        }
        while(pointer1 != pointer2)
        {
            int area = (Math.min(A[pointer2], A[pointer1]) * (pointer2 - pointer1));
            //System.out.println(area);
            if(area > maximum)
            {
                maximum = area;
            }
            if(A[pointer2] < A[pointer1])
            {
                pointer2--;
            }
            else
            {
                pointer1++;
            }
        }
        return maximum;
    }
}