container with most water sliding window solution

Guri Bee

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19-10-2024

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Cracking the Container with Most Water: A Sliding Window Approach

The "Container with Most Water" problem is a classic coding challenge that tests your ability to think algorithmically. It asks you to find the container that can hold the most water given an array of non-negative integers representing the height of each container's vertical walls.

Let's dive into the solution using the efficient sliding window technique, inspired by insights from GitHub discussions.

Understanding the Problem

Imagine a series of vertical lines arranged in a horizontal row, with each line's height represented by an integer in the input array. These lines form containers with the x-axis as their base. The objective is to find the container with the maximum area.

Key Idea: The area of a container is determined by the shorter of its two vertical lines (height) and the distance between them (width).

The Sliding Window Technique

The sliding window approach offers a clever and efficient solution. Instead of checking all possible container pairs, we iteratively move two pointers (left and right) towards each other, continuously updating the maximum area:

1. Initialization: Start with two pointers, left at index 0 and right at the end of the array.
2. Iteration:
   • Calculate the current area using min(height[left], height[right]) * (right - left).
   • Update the maxArea if the current area is larger.
   • Move the pointer with the shorter height: If height[left] < height[right], move left one step to the right. Otherwise, move right one step to the left.
3. Termination: Continue until left and right pointers cross each other.

Code Implementation (Python)

def maxArea(height):
    """
    Finds the container with the maximum area.

    Args:
        height: An array of non-negative integers representing the height of each container's vertical walls.

    Returns:
        The maximum area of a container.
    """
    left = 0
    right = len(height) - 1
    maxArea = 0

    while left < right:
        currentArea = min(height[left], height[right]) * (right - left)
        maxArea = max(maxArea, currentArea)

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

    return maxArea

# Example usage
height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
max_area = maxArea(height)
print(f"Maximum area: {max_area}")

Explanation:

• The while loop iterates until the pointers cross each other, ensuring all possible container combinations are considered.
• The min(height[left], height[right]) line determines the height of the current container.
• The right - left calculation provides the width of the container.
• The max(maxArea, currentArea) line updates the maximum area found so far.
• The pointer movement logic guarantees that the maximum area can only increase or remain the same, making the algorithm efficient.

Understanding the Optimization

This solution utilizes the sliding window concept by narrowing the search space with each iteration. The core optimization lies in the logic of moving the pointer associated with the shorter height.

Why move the shorter pointer? By moving the shorter pointer, we ensure the following:

• Potential for increased area: Moving the shorter pointer can potentially increase the width of the container, thus increasing the area if the taller line remains fixed.
• Eliminating redundant checks: Moving the shorter pointer eliminates the need to check containers formed by the current shorter line with lines to its left. This is because any container formed with a line to the left would be smaller, as it would have the same or smaller height but a smaller width.

Conclusion

The sliding window approach elegantly solves the "Container with Most Water" problem. This method is highly efficient with a time complexity of O(n), where n is the length of the input array. By strategically moving the pointers and focusing on the shorter line, we effectively prune the search space and arrive at the optimal solution.