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
Container With Most Water Solutions
✅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
Watch Tutorial
Checkout more Solutions here