The O(n) constraint is what makes this worth thinking about.
Start by clarifying the problem and confirming that the container's area is determined by the shorter line and the distance between lines. Then explain the two-pointer technique: initialize pointers at both ends, compute the area, and move the pointer pointing to the shorter line inward, repeating until pointers meet. Emphasize that this yields O(n) time and O(1) space.
Pro tip: During the explanation, explicitly state why moving the shorter line is safe: any container formed with the shorter line and any other line cannot exceed the current area because the height is limited by the shorter line and the width will be smaller. This demonstrates deep understanding and avoids the common pitfall of moving the taller line.
Restate the problem to ensure understanding: given an array of heights, find two lines that together with the x-axis form a container holding the most water. Confirm that the container cannot be slanted and that the area is min(height[i], height[j]) * (j - i).
Mention that a brute force approach would check all pairs in O(n^2) time, which is inefficient for large inputs. This sets the stage for the need for an O(n) solution.
Explain that we can use two pointers, one at the beginning and one at the end of the array. The area is computed as the minimum of the two heights multiplied by the distance between them.
After computing the area, move the pointer that points to the shorter line inward. Justify this by noting that moving the taller line cannot increase the area because the height is limited by the shorter line and the width decreases.
Conclude that the algorithm runs in O(n) time and O(1) space. Mention edge cases such as arrays with fewer than two elements or all equal heights, and confirm the algorithm handles them correctly.
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