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.
1#include <iostream>
2#include <vector>
3
4int maxArea(std::vector<int>& height) {
5 int left = 0, right = height.size() - 1;
6 int max_area = 0;
7 while (left < right) {
8 int h = std::min(height[left], height[right]);
9 max_area = std::max(max_area, h * (right - left));
10 height[left] < height[right] ? left++ : right--;
11 }
12 return max_area;
13}
14
15int main() {
16 std::vector<int> height = {1, 8, 6, 2, 5, 4, 8, 3, 7};
17 std::cout << "Max area: " << maxArea(height) << std::endl;
18 return 0;
19}The C++ implementation uses a std::vector to store the heights and applies the two-pointer technique in a similar manner to maximize the water container area.
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.
1class Solution:
2 def maxArea(self, height):
3 max_area = 0
4 for i in range(len(height) - 1):
5 for j in range(i + 1, len(height)):
6 h = min(height[i], height[j])
7 max_area = max(max_area, h * (j - i))
8 return max_area
9
10
11if __name__ == "__main__":
12 sol = Solution()
13 print(sol.maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7]))This brute force Python method calculates the maximum container area by evaluating all pairs of lines' possible areas and selecting the maximum.