content: package codingquestions.leetcode;
  
  /*
   * 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
   * 
   */
  public class ContainerMostWater {
  
      public static void main(String[] args) {
          var cmw = new ContainerMostWater();
          System.out.println(cmw.maxArea(new int[]{1, 8, 6, 2, 5, 4, 8, 3, 7}));
          ;
          System.out.println(cmw.maxAremyMethod(new int[]{1, 8, 6, 2, 5, 4, 8, 3, 7}));
          ;
      }
  
      public int maxAremyMethod(int[] height) {
          int area = Integer.MIN_VALUE;
  
          for (int i = 0; i < height.length; i++) {
              for (int j = i; j < height.length; j++) {
                  int areas = (j - i) * (Math.min(height[i], height[j]));
                  if (areas > area) {
                      area = areas;
                  }
              }
          }
          return area;
      }
  
      public int maxArea(int[] height) {
          int size = height.length;
          if (size == 1) {
              return 0;
          }
          if (size == 2) {
              return Math.min(height[0], height[1]);
          }
  
          int left = 0;
          int right = size - 1;
          int max_area = 0;
  
          while (left != right) {
              if (height[left] < height[right]) {
                  max_area = Math.max(max_area, height[left] * (right - left));
                  left++;
              } else {
                  max_area = Math.max(max_area, height[right] * (right - left));
                  right--;
              }
          }
          return max_area;
  
      }
  }
  