Container With Most Water 2 Pointer Approach

 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

 

🧠 2. Two Pointer Approach (Simple Idea)

1. Start with two pointers:

   • Left pointer at start (i = 0)

   • Right pointer at end (j = height.length - 1)

2. While i < j:

   • Calculate the current water area.

   • Keep track of the maximum area found.

   • Move the pointer with shorter height, because:

     • Only moving the smaller line may increase the height (and improve area).

class Solution {

    public int maxArea(int[] height) {

       

        int i = 0, j = height.length - 1;

        int maxArea = 0;

        while (i < j) {

            int h = Math.min(height[i], height[j]);

            int w = j - i;

            int area = h * w;

            maxArea = Math.max(maxArea, area);

            // Move the smaller height pointer

            if (height[i] < height[j]) {

                i++;

            } else {

                j--;

            }

        }

        return maxArea;

    }

}