The two pointers approach efficiently searches for the maximum area by starting with the widest possible container and gradually narrowing it:
This approach works due to the observation that the area is limited by the shorter line, so the only way to get a larger area is to find a taller line.
Time Complexity: O(n), where n is the number of elements in the height array, due to the single traversal of the array.
Space Complexity: O(1) as only a few extra variables are used.
1public class Solution {
2 public int maxArea(int[] height) {
3 int left = 0, right = height.length - 1;
4 int maxArea = 0;
5 while (left < right) {
6 int h = Math.min(height[left], height[right]);
7 maxArea = Math.max(maxArea, h * (right - left));
8 if (height[left] < height[right]) {
9 left++;
10 } else {
11 right--;
12 }
13 }
14 return maxArea;
15 }
16
17 public static void main(String[] args) {
18 Solution solution = new Solution();
19 int[] height = {1,8,6,2,5,4,8,3,7};
20 System.out.println("Max area: " + solution.maxArea(height));
21 }
22}In this Java solution, the two-pointer technique iteratively calculates and updates the maximum possible water container area. We use the length property of the array for the right pointer.
Although not optimal for large inputs, the brute force approach explores every possible pair of lines to find the maximum container area:
However, the complexity of this approach makes it unsuitable for large datasets due to its quadratic time complexity.
Time Complexity: O(n^2), where n is the number of elements as each pair is checked.
Space Complexity: O(1) due to a linear amount of extra space.
1using System;
2
3public class Solution
4{
5 public int MaxArea(int[] height)
6 {
7 int maxArea = 0;
8 for (int i = 0; i < height.Length - 1; i++)
9 {
10 for (int j = i + 1; j < height.Length; j++)
11 {
12 int h = Math.Min(height[i], height[j]);
13 maxArea = Math.Max(maxArea, h * (j - i));
14 }
15 }
16
17 return maxArea;
18 }
19
20 public static void Main(string[] args)
21 {
22 Solution solution = new Solution();
23 int[] height = {1,8,6,2,5,4,8,3,7};
24 Console.WriteLine("Max area: " + solution.MaxArea(height));
25 }
26}The C# brute force solution uses nested loops to check all possible pairs of lines and computes the container's maximum water area.