# Container With Most Water Leetcode Solution | Easy Approach Share:

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`

Time: O(1)
Space: O(1)

### 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;
}
};

``````

### 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;
}
}

``````

### 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

``````

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