Package g0001_0100.s0011_container_with_most_water

Class Solution

java.lang.Object
g0001_0100.s0011_container_with_most_water.Solution
public class Solution extends Object
11 - Container With Most Water.

Medium

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 the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

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

Example 3:

Input: height = [4,3,2,1,4]

Output: 16

Example 4:

Input: height = [1,2,1]

Output: 2

Constraints:

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

To solve the Container With Most Water problem in Java using a Solution class, we’ll follow these steps:

  1. Define a Solution class with a method named maxArea that takes an array of integers height as input and returns the maximum area of water that can be contained.
  2. Initialize two pointers, left pointing to the start of the array and right pointing to the end of the array.
  3. Initialize a variable maxArea to store the maximum area encountered so far, initially set to 0.
  4. Iterate while left is less than right.
  5. Calculate the current area using the formula: (right - left) * min(height[left], height[right]).
  6. Update maxArea if the current area is greater than maxArea.
  7. Move the pointer pointing to the smaller height towards the other pointer. If height[left] < height[right], increment left, otherwise decrement right.
  8. Continue the iteration until left becomes greater than or equal to right.
  9. Return maxArea.

Here’s the implementation:

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

        while (left < right) {
            int currentArea = (right - left) * Math.min(height[left], height[right]);
            maxArea = Math.max(maxArea, currentArea);

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

        return maxArea;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();

        // Test cases
        int[] height1 = {1, 8, 6, 2, 5, 4, 8, 3, 7};
        System.out.println("Example 1 Output: " + solution.maxArea(height1));

        int[] height2 = {1, 1};
        System.out.println("Example 2 Output: " + solution.maxArea(height2));

        int[] height3 = {4, 3, 2, 1, 4};
        System.out.println("Example 3 Output: " + solution.maxArea(height3));

        int[] height4 = {1, 2, 1};
        System.out.println("Example 4 Output: " + solution.maxArea(height4));
    }
}

This implementation provides a solution to the Container With Most Water problem in Java.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maxArea

      public int maxArea(int[] height)