11. Container With Most Water
MediumLeetCodeYou 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 <= 10^5
0 <= height[i] <= 10^4
How to solve the problem
- Assuming we have
height[left]
from start,height[right]
from end, andheight[left]
>height[right]
. No matter how we changeheight[left]
, the biggest area is always equal toheight[left]
*height[right]
, because the area is determined by the short height which isheight[right]
. So, we need to change the short height toright--
.
Code
- Two Pointer
Python
class Solution:
def maxArea(self, height: List[int]) -> int:
left = 0
right = len(height) - 1
maxArea = 0
while left < right:
area = (right - left) * min(height[left], height[right])
maxArea = max(maxArea, area)
if height[left] > height[right]:
right -= 1
else:
left += 1
return maxArea
- Brute Force
Python
class Solution:
def maxArea(self, height: List[int]) -> int:
left = 0
right = len(height)
maxArea = 0
for left in range(len(height)):
for right in range(left+1, len(height)):
area = min(height[left], height[right]) * (right - left)
maxArea = max(area, maxArea)
return maxArea
Complexity
Time complexity:
- Two Pointer: O(n)
- Brute Force: O(n^2)
Space complexity: O(1)
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