Container With Most Water – Solution & Complexity

Q: What is the time and space complexity of Container With Most Water?

The optimal Container With Most Water solution runs in `O(n)` time and `O(1)` space. Two pointers from both ends, always moving the shorter line inward.

Q: What is the most efficient way to solve Container With Most Water?

Two pointers from both ends, always moving the shorter line inward. This gives `O(n)` time complexity, and is a classic arrays, two pointers, and greedy problem.

Q: How do you solve Container With Most Water in Python & JavaScript?

We provide a worked Container With Most Water solution in Python & JavaScript with a step-by-step explanation above. You can also open the interactive editor to write and run your own Python solution against the test cases in the browser.

Q: Is Container With Most Water an easy or hard interview question?

Container With Most Water is generally rated **Medium**. It tests arrays, two pointers, and greedy, which are common coding-interview patterns.

1. Understand the Area Formula

Choose two vertical lines at indices left and right.
The width is right - left.
The height of the water is limited by the shorter line.
Area is (right - left) * min(height[left], height[right]).

2. Brute Force Idea

Check every pair of lines and compute its area.
Keep the maximum area found.
This is simple but O(n²), which is too slow for large arrays.

def max_area(height):
    best = 0
    for left in range(len(height)):
        for right in range(left + 1, len(height)):
            width = right - left
            water_height = min(height[left], height[right])
            best = max(best, width * water_height)
    return best

3. Use Two Pointers

Start with the widest possible container: first and last lines.
Compute the area, then move one pointer inward.
Moving the taller line cannot help because the shorter line still limits the height while width shrinks.
Therefore, always move the pointer at the shorter line.

4. Update the Best Area While Moving

At each step, calculate the current area.
Update best if this area is larger.
Move left when the left line is shorter, otherwise move right.

left = 0
right = len(height) - 1
best = 0

while left < right:
    width = right - left
    water_height = min(height[left], height[right])
    best = max(best, width * water_height)

    if height[left] < height[right]:
        left += 1
    else:
        right -= 1

5. Final Solution and Complexity

The two pointers scan inward in one pass.
Every possible useful candidate is considered without checking all pairs.
Time complexity is O(n).
Space complexity is O(1).

def max_area(height: list[int]) -> int:
    left = 0
    right = len(height) - 1
    best = 0

    while left < right:
        width = right - left
        water_height = min(height[left], height[right])
        best = max(best, width * water_height)

        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return best

Container With Most Water – Solution & Complexity

1. Understand the Area Formula

2. Brute Force Idea

3. Use Two Pointers

4. Update the Best Area While Moving

5. Final Solution and Complexity

FAQ

What is the time and space complexity of Container With Most Water?

What is the most efficient way to solve Container With Most Water?

How do you solve Container With Most Water in Python & JavaScript?

Is Container With Most Water an easy or hard interview question?