**Problem Statement:**

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

**Solution:**

**Algorithm:**

- Initialize the maxArea to 0
- Initialize the left pointer to 0 and the right pointer to the length of the array - 1
- Loop through the array while the left pointer is less than the right pointer
- Calculate the area by multiplying the minimum of the height of the left and right pointer by the difference of the right and left pointer
- Update the maxArea by comparing the maxArea and the area
- If the height of the left pointer is less than the height of the right pointer, increment the left pointer, else decrement the right pointer
- Return the maxArea

**Code:**

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

**Time Complexity:**

*O(n)*

**Space Complexity:**

*O(1)*

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