You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the i^{th} 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 <= 10^{5}

0 <= height[i] <= 10^{4}

Container With Most Water LeetCode Solution in C++

int maxArea(vector<int>& height) {
int water = 0;
int i = 0, j = height.size() - 1;
while (i < j) {
int h = min(height[i], height[j]);
water = max(water, (j - i) * h);
while (height[i] <= h && i < j) i++;
while (height[j] <= h && i < j) j--;
}
return water;
}

Container With Most Water LeetCode Solution in Java

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

Container With Most Water LeetCode Solution in Python

class Solution:
def maxArea(self, H: List[int]) -> int:
ans, i, j = 0, 0, len(H)-1
while (i < j):
if H[i] <= H[j]:
res = H[i] * (j - i)
i += 1
else:
res = H[j] * (j - i)
j -= 1
if res > ans: ans = res
return ans

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