# Container With Most Water – Leetcode Solution

In this post, we are going to solve the 11. Container With Most Water problem of Leetcode. This problem 11. Container With Most Water is a Leetcode medium level problem. Let’s see code, 11. Container With Most Water.

## Problem

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`

Now, let’s see the code of 11. Container With Most Water – Leetcode Solution.

# Container With Most Water – Leetcode Solution

### 11. Container With Most Water – Solution in Java

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

### 11. Container With Most Water – Solution in C++

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

### 11. Container With Most Water– Solution in Python

```class Solution:
def maxArea(self, height: List[int]) -> int:
p1 = 0
p2 = len(height) - 1
max_area = 0
while p1 != p2:
if height[p1] > height[p2]:
area = height[p2] * (p2 - p1)
p2 -= 1
else:
area = height[p1] * (p2 - p1)
p1 += 1
if area > max_area: max_area = area
return max_area```

Note: This problem 11. Container With Most Water is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.