## Algorithm

Problem Name: 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`

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
int maxArea(int* height, int heightSize) {
int l, r, x, water, max = 0;
int i, j;
#if 0
for (i = 0; i < heightSize - 1; i ++) {
if (i == 0 || height[i] > l) {
l = height[i];
for (j = i + 1; j < heightSize; j ++) {
r = height[j];
x = l < r ? l : r;
water = x * (j - i);
if (max < water) max = water;
}
}
}
#else
i = 0;
j = heightSize - 1;
while (i < j) {
l = height[i];
r = height[j];
x = l < r ? l : r;
water = x * (j - i);
if (max < water) max = water;
if (l < r) i ++;
else j --;
}
#endif
return max;
}
``````
Copy The Code &

Input

cmd
height = [1,8,6,2,5,4,8,3,7]

Output

cmd
49

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
int maxArea(vector& height) {
int maxArea = 0, minLeft = 0, minRight = 0;
for(int i = height.size() - 1; i >= 0; i--){
if(height[i] < minRight) continue;
minLeft = 0;
for(int j = 0; j < i; j++){
if(height[j] < minLeft) continue;
maxArea = max(maxArea, min(height[i], height[j]) * (i - j));
minLeft = max(minLeft, height[j]);
}
minRight = max(minRight, height[i]);
}
return maxArea;
}
};
``````
Copy The Code &

Input

cmd
height = [1,1]

Output

cmd
1

### #3 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int maxArea(int[] height) {
int maximumArea = 0;
int leftIdx = 0;
int rightIdx = height.length - 1;
while (leftIdx < rightIdx) {
int maxHeight = Math.min(height[leftIdx], height[rightIdx]);
int currArea = maxHeight * (rightIdx - leftIdx);
maximumArea = Math.max(currArea, maximumArea);
if (maxHeight == height[leftIdx]) {
leftIdx++;
} else {
rightIdx--;
}
}
return maximumArea;
}
}
``````
Copy The Code &

Input

cmd
height = [1,1]

Output

cmd
1

### #4 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const maxArea = function(height) {
let res = 0, l = 0, r = height.length - 1
while(l < r) {
const tmp = (r - l) * Math.min(height[l], height[r])
if(tmp > res) res = tmp
if(height[l] < height[r]) l++
else r--
}
return res
};

``````
Copy The Code &

Input

cmd
height = [1,8,6,2,5,4,8,3,7]

Output

cmd
49