11.Container_With_Most_Water(Medium).md

11. Container With Most Water (Easy)

Date and Time: Jul 8, 2024, 23:56 (EST)

Link: https://leetcode.com/problems/container-with-most-water/

Question:

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

KeyPoints:

This question is similar to [42. Trapping Rain Water]. Using the two pointers method, we calculate the area by min(height[l], height[r]) * (r - l), then we move the min(height[l], height[r]) because we want to get a greater area by taking larger height.

My Solution:

Apr 19, 2026 11:08:00 AM (PDT), Time Taken: 13m 59s

class Solution:
    def maxArea(self, height: List[int]) -> int:
        # Return the max area from two heights
        # Two ptrs, maxLeft and maxRight, greedily maximize the area
        # 1. Update maxLeft if maxLeft <= maxRight, and update the area (max(current area, currentMax))
        # 2. Update maxRight, when maxLeft > maxRight, update maxRight
        # 3. area = min(maxLeft, maxRight) * (r - l)
        # TC: O(n), n=len(height), SC: O(1)
        l, r = 0, len(height) - 1
        area = 0
        while l < r:
            maxL, maxR = height[l], height[r]
            area = max(area, min(maxL, maxR) * (r - l))
            if maxL <= maxR:
                l += 1
            else:
                r -= 1
        return area

Time Complexity: $O(n)$
Space Complexity: $O(1)$