Container With Most Water

Place one pointer at each end of the height array. The area is (right - left) * min(height[left], height[right]). Record the best area, then move the pointer at the shorter line inward.

The move is justified by elimination. Suppose the left line is shorter. Any container that keeps that left line and moves the right pointer inward has a smaller width and its height is still limited by the same left line, so it cannot beat the current area. The only possibility of improvement is to discard the shorter line and search for a taller boundary. Apply the symmetric rule when the right line is shorter; when equal, either side may move.

Each pointer moves inward at most n - 1 times, so time is O(n) and extra space is O(1). A brute-force comparison of every pair is O(n^2). Use a sufficiently wide numeric type for the multiplication when constraints permit large heights or lengths.