11. 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.

Example 1:

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.

Example 2:

Input: height = [1,1]
Output: 1

Constraints:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

Solution:

class Solution {
    public int maxArea(int[] height) {
        int result = 0;

        /*
             0 1 2 3 4 5 6 7 8
            [1,8,6,2,5,4,8,3,7]
               i 
                - - - - - - - 
                             j

            lmax = 8 
            rmax = 7 
            area = Math.min(lmax, rmax) * (8-1)
                 =  7 *  7 = 49

        */

        int left = 0;
        int lmax = height[left];

        int right = height.length - 1;
        int rmax = height[right];

        while(left < right){
            int area = Math.min(lmax, rmax) * (right - left);
            result = Math.max(area, result);

            int curLeft = height[left];
            int curRight = height[right];

            if (curLeft <= curRight){
                left++;
                lmax = Math.max(height[left], lmax);
            }else{
                right--;
                rmax = Math.max(height[right], rmax);
            }


        }

        return result;
    }
}

// TC: O(n)

// SC: O(1)

class Solution {
    public int maxArea(int[] height) {
        int right = height.length - 1;
        int left = 0;

        int max = 0;
        while(left < right){
            int width = right - left;

            int h = Math.min(height[left], height[right]);

            max = Math.max(width * h, max);

            if (height[left] <= height[right]){
                left++;
            }else{
                right--;
            }

        }

        return max;

    }
}

// TC: O(n)
// SC: O(1)

/* 

 [1,8,6,2,5,4,8,3,7]
    l             r


*/