11. Container With Most Water — LeetCode(Python)
I got you!
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: 1Constraints:
n == height.length2 <= n <= 1050 <= height[i] <= 104
Solution:
l, r = 0, len(height) - 1
water = 0
while l < r:
water = max(water, min(height[l], height[r]) * (r - l))
if height[l] < height[r]:
l += 1
else:
r -= 1
return waterExplanation —
We use a two-pointer approach here.
We initialize two pointers l and r and point them at two ends of the array.
The water contained by a container is the product of it’s width, r-l, and the height of it’s shorter boundary wall, min(height[l], height[r]).
This concept is used to keep updating the maximum amount of water contained by any container in the variable water.
We increment our left pointer if the height[l] < height[r], and decrement our right pointer otherwise. Note that shifting any pointer is fine if height[l] == height[r].
Finally , we return our water variable.
Time and Space Complexity —
Since we visit each element in the array nums only once, we can say that our code runs in linear time. Also, we require constant extra space.
Thus, if n is the length of the array nums,
Time Complexity: O(n)
Space Complexity: O(1)
