11. Container With Most Water

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

Constraints:

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

Approach 01: Two Pointers

class Solution {
public:
    int maxArea(vector& height) {
        int left = 0, right = height.size() - 1;
        int max_area = 0;
        while (left < right) {
            int h = min(height[left], height[right]);
            max_area = max(max_area, h * (right - left));
            if (height[left] < height[right]) left++;
            else right--;
        }
        return max_area;
    }
};

class Solution:
    def maxArea(self, height: List[int]) -> int:
        left, right = 0, len(height) - 1
        max_area = 0
        while left < right:
            h = min(height[left], height[right])
            max_area = max(max_area, h * (right - left))
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1
        return max_area

class Solution {
    public int maxArea(int[] height) {
        int left = 0, right = height.length - 1;
        int maxArea = 0;
        while (left < right) {
            int h = Math.min(height[left], height[right]);
            maxArea = Math.max(maxArea, h * (right - left));
            if (height[left] < height[right]) left++;
            else right--;
        }
        return maxArea;
    }
}