What is the best approach to solve Container With Most Water?

The optimal approach for Container With Most Water is the "Optimal — Two Pointers" technique. It runs in O(n) time and uses O(1) space. Start with the widest container (lo=0, hi=n-1). The area is limited by the shorter line. Moving the taller pointer inward can only decrease width without increasing height — so always move the shorter

What is the time complexity of Container With Most Water?

The optimal solution for Container With Most Water has a time complexity of O(n) and a space complexity of O(1), achieved using the Two Pointers pattern.

Is Container With Most Water a Medium LeetCode problem?

Yes, LeetCode #11 (Container With Most Water) is rated Medium. It uses the Two Pointers pattern and is commonly asked in interviews at Amazon, Microsoft, Goldman Sachs.

How to solve Container With Most Water in Python, Java, C++, and JavaScript?

LumiChats provides complete solutions for Container With Most Water in Python, Java, C++, and JavaScript, along with step-by-step visual animations and time/space complexity analysis. All solutions are free to view.

#11MediumTwo Pointers

Container With Most Water

Shrink from both ends, always moving the shorter height inward.

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Problem Statement

You are given an integer array height of length n. There are n vertical lines where the endpoints of line i are (i, 0) and (i, height[i]). Find two lines that together with the x-axis form a container that holds the most water.

Examples

Example 1

Input:height = [1,8,6,2,5,4,8,3,7]

Output:49

Explanation: Lines at index 1(h=8) and 8(h=7): width=7, min(8,7)=7, area=49.

Example 2

Input:height = [1,1]

Output:1

Constraints

n == height.length
2 ≤ n ≤ 10⁵
0 ≤ height[i] ≤ 10⁴

Solutions

Brute Force

TimeO(n²)SpaceO(1)

Try every pair of lines and compute water = (j - i) * min(height[i], height[j]). O(n²).

Visual Animation

def maxArea(height: list[int]) -> int:
    max_water = 0
    n = len(height)
    for i in range(n):
        for j in range(i+1, n):
            water = (j - i) * min(height[i], height[j])
            max_water = max(max_water, water)
    return max_water

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