LeetCode 11 Solution: Container with Most Water

Today, we’ll explore LeetCode Solution 11: Container with Most Water, a classic problem that highlights the importance of efficient problem-solving techniques. If you’re gearing up for coding interviews or sharpening your algorithm skills, this problem is a great opportunity to practice optimizing solutions using the two-pointer technique.

Problem Overview: Container with Most Water

The task in this problem is simple yet intriguing. You are given an array of integers called height, where each element represents the height of a vertical line drawn at that index. The goal is to find two lines that, along with the x-axis, can form a container capable of holding the maximum amount of water. The trick, however, lies in maximizing the area formed by these two lines.

The area of water the container can hold is determined by two factors: the distance between the lines (the width), and the height of the container, which is dictated by the shorter of the two lines. Mathematically, the area can be calculated as:

Area = width x min(height[left], height[right])

Where left and right are the indices of the two lines being considered, and the width is simply the difference between these indices. The key challenge is to maximize this area by adjusting which two lines are chosen, but doing so efficiently.

Constraints to Consider

The problem can involve up to 10⁵ lines, so a brute-force solution that checks every possible pair of lines would be far too slow. A brute-force method would take (O(n²)) time, which would be inefficient given the input size. Therefore, we need a solution that operates in linear time, (O(n)). Each line’s height can be as high as 10,000 or as low as 0, so our solution must also account for edge cases like zero-height lines.

The Two-Pointer Approach to LeetCode 11 Solution

To solve this problem efficiently, we can employ the two-pointer technique. Instead of comparing every possible pair of lines, we start by placing one pointer at the beginning of the array and the other at the end. We then calculate the area formed by these two lines and gradually adjust the pointers to explore better options. The goal is to move the pointer that is pointing to the shorter line inward, hoping that this will lead to a taller line and a potentially larger container.

By calculating the area with the current two pointers and adjusting the pointers intelligently, we avoid unnecessary comparisons. The reason this works is that the height of the water is always determined by the shorter of the two lines. Thus, moving the pointer at the shorter line might give us a chance to find a taller line and, hopefully, a larger area.

Python Code Implementation

Here’s how we can implement this two-pointer approach in Python:

def maxArea(height):
    left = 0
    right = len(height) - 1
    max_area = 0

    while left < right:
        current_area = min(height[left], height[right]) * (right - left)
        max_area = max(max_area, current_area)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return max_area

This code begins by initializing two pointers, one at the start of the array and one at the end. It then enters a loop where, at each iteration, the area formed by the two pointers is calculated, and the pointer that points to the shorter line is moved inward. The process continues until the two pointers meet in the middle, ensuring that all potential container configurations are explored efficiently. Throughout this, we keep track of the maximum area encountered and return it as the final result.

Exploring the Example Walkthrough

Consider the example where height = [1, 8, 6, 2, 5, 4, 8, 3, 7]. In this case, the two lines that form the container with the most water are at indices 1 and 8, with heights of 8 and 7. The width between them is 7, so the area is calculated as ( 7 times 7 = 49 ). This is the maximum possible area for this input.

In another scenario where height = [1, 1], the lines both have a height of 1, and the distance between them is 1. As a result, the area is simply ( 1 times 1 = 1 ).

Handling Edge Cases

Edge cases also play a role in this problem. If all the heights are zero, no water can be contained, resulting in an area of zero. Similarly, if the height of all lines is the same, the container with the largest width will yield the maximum area. For example, if height = [1, 1, 1, 1], the area will be determined by the distance between the first and last line.

Time and Space Complexity

The time complexity of this algorithm is (O(n)), where (n) is the number of lines. This is because the two-pointer technique allows us to make a single pass through the array. The space complexity is (O(1)), as we only need a few extra variables (left, right, max_area) to store intermediate values during the calculation.

Conclusion

LeetCode Solution 11: Container with Most Water is an excellent example of how to apply the two-pointer technique to optimize a problem that initially seems to require a brute-force solution. By cleverly adjusting which pointers to move based on the height of the lines, we can efficiently maximize the container’s area in linear time.

This problem is frequently encountered in coding interviews, so mastering it will give you valuable insight into how to solve similar optimization problems with two pointers. Keep practicing, and soon enough, you’ll be able to apply this technique with confidence in a variety of contexts.

Happy coding!

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Looking for more LeetCode solutions? Stay tuned for upcoming guides on tackling other LeetCode challenges in Python. Whether you’re preparing for technical interviews or aiming to improve your coding skills, exploring a variety of problems will enhance your understanding and proficiency.

You can find this problem on LeedCode, click here.