How would you solve the 'Container With Most Water' problem: find two lines that together with the x-axis form a container holding the maximum water?

Overview

This interview question is a staple in technical screens and onsite interviews, particularly at FAANG companies. It directly tests a candidate's ability to move beyond naive solutions and identify optimal algorithms, demonstrating a strong grasp of data structures, algorithmic efficiency, and clean code practices essential for building high-performance software.

Interview Question:

How would you solve the 'Container With Most Water' problem: find two lines that together with the x-axis form a container holding the maximum water?

Why Interviewers Ask This:

Interviewers pose this problem to evaluate several critical aspects of a Software Engineer's skill set. First, they want to assess your problem-solving methodology and ability to break down a seemingly complex problem into manageable parts. More importantly, it gauges your understanding of algorithmic optimality and your capacity to identify and implement solutions with the best possible time and space complexity. They are looking for candidates who can articulate the trade-offs between different approaches, such as a brute-force O(N^2) solution versus an optimized O(N) solution. Finally, it provides insight into your ability to write clean, production-grade code that is readable, maintainable, and handles edge cases gracefully, reflecting your commitment to software design principles.

Expert Answer:

To solve the Container With Most Water problem optimally, I would employ a two-pointer approach. First, I would initialize two pointers, one at the beginning of the array (left pointer) and one at the end of the array (right pointer). I would also maintain a variable to store the max_area found so far, initialized to zero. The core idea is that the area of a container is determined by the shorter of the two lines multiplied by the distance between them.

Next, I would enter a loop that continues as long as the left pointer is less than the right pointer. Inside the loop, I would calculate the current area. The height of the container is the minimum of the heights at the left and right pointers, and the width is the difference between their indices (right - left). I would then update max_area if the current_area is greater. This step ensures we always keep track of the largest container found.

Then, the crucial decision is how to move the pointers. To potentially find a larger area, we must try to increase the height of the container. Since the area is limited by the shorter line, moving the taller line inward will never increase the height, and might even decrease the width. Therefore, we should always move the pointer corresponding to the shorter line inward. If height[left] is less than height[right], I would increment the left pointer. Otherwise, I would decrement the right pointer. This strategy is key to the algorithm's efficiency, as it systematically explores potential larger containers.

Finally, once the left pointer crosses or meets the right pointer, the loop terminates, and the max_area variable will hold the maximum amount of water that can be contained. This approach guarantees an O(N) time complexity because each pointer traverses the array at most once. The space complexity is O(1) as we only use a few variables, making it an extremely efficient and production-ready solution suitable for large datasets.

Speaking Blueprint:

[The Hook] This is a classic problem that tests our ability to find an optimal solution. My approach would leverage a two-pointer technique, which is highly efficient for this type of problem, achieving linear time complexity.

[The Core Execution] I would initialize two pointers, one at each end of the array, and a variable to track the maximum area. In each step, I calculate the area formed by the current lines, taking the minimum of their heights and multiplying by their distance. The critical part is deciding which pointer to move: I always advance the pointer associated with the shorter line. This is because moving the taller line inward cannot increase the container's height and only reduces its width, whereas moving the shorter line offers the potential to find a taller boundary and thus a larger area. This process continues until the pointers meet.

[The Punchline] This two-pointer strategy ensures we explore all relevant configurations optimally. It provides an O(N) time complexity and O(1) space complexity, making it a robust and scalable solution for real-world applications where performance is critical. I am confident this approach is both correct and highly efficient.

Common Mistakes:

• Brute-force approach: Implementing an O(N^2) solution by checking every possible pair of lines. While correct, it fails to meet the optimality expectations for a senior software engineer role, especially in a FAANG interview setting.
• Incorrect pointer movement: Not understanding why only the shorter pointer should be moved. Moving the taller pointer or both arbitrarily can lead to missing the optimal solution or an inefficient algorithm.
• Off-by-one errors: Miscalculating the width of the container (e.g., using right - left + 1 instead of right - left) or incorrect loop termination conditions.
• Ignoring edge cases: Not considering arrays with fewer than two elements, or arrays where all elements are the same height, or arrays with zero-height lines. A robust solution must handle these gracefully.
• Lack of Big O analysis: Failing to articulate the time and space complexity of the proposed solution. This is a critical component of demonstrating algorithmic understanding and is expected for optimal solutions.