Intuition

• Given 𝑛 n non-negative integers 𝑎 1 , 𝑎 2 , … , 𝑎 𝑛 a 1 ​ ,a 2 ​ ,…,a n ​ where each integer represents the height of a vertical line drawn at that point on the x-axis, the task is to find two lines that, together with the x-axis, form a container that holds the most water.
• The amount of water container by any two lines is determined by multiplying the shorter of the two lines and the distance between them.

Approach

1. Initial two pointers

• Start with the pointer at the beginning (start) and one at the end (end) of the array

2.Calculate the area

• Compute the area formed by the lines these two pointers using formula

          Area = min(a[start], a[end]) * (end - start)
  

  3. Move Pointers:

  • To potential find a larger area, move the pointer pointing to the short line inward

  4. Track the Maximum Length

  • Keep track of the maximum area found during the iterations.

  Complexity

• Time complexity: O(N)

• Space complexity: O(1).

Code

class Solution {
    public int maxArea(int[] height) {
        int start = 0;
        int end = height.length - 1;
        int max = 0;

        while (start < end) {
            if (height[start] < height[end]) {
                max = Math.max(max, height[start] * (end - start));
                start++;
            } else {
                max = Math.max(max, height[end] * (end - start));
                end--;
            }
        }

        return max;
    }
}