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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 `i`

line are ^{th}`(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:49Explanation: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 <= 10`

^{5}`0 <= height[i] <= 10`

^{4}

**Time:**O(1)*O*(1)**Space:**O(1)*O*(1)

### Container With Most Water Program Solution C++

```
class Solution {
public:
int maxArea(vector<int>& height) {
int ans = 0;
int l = 0;
int r = height.size() - 1;
while (l < r) {
const int minHeight = min(height[l], height[r]);
ans = max(ans, minHeight * (r - l));
if (height[l] < height[r])
++l;
else
--r;
}
return ans;
}
};
```

### Container With Most Water Program Solution JAVA

```
class Solution {
public int maxArea(int[] height) {
int ans = 0;
int l = 0;
int r = height.length - 1;
while (l < r) {
final int minHeight = Math.min(height[l], height[r]);
ans = Math.max(ans, minHeight * (r - l));
if (height[l] < height[r])
++l;
else
--r;
}
return ans;
}
}
```

### Container With Most Water Program Solution PYTHON

```
class Solution:
def maxArea(self, height: List[int]) -> int:
ans = 0
l = 0
r = len(height) - 1
while l < r:
minHeight = min(height[l], height[r])
ans = max(ans, minHeight * (r - l))
if height[l] < height[r]:
l += 1
else:
r -= 1
return ans
```