technique overview
++ container with most water +
+Sometimes, the mathematical solution is the simplest.
++ The area of a container bounded by the ground and its columns at + positions \((l, r)\) is: \[ \text{area} = \text{width} \cdot + \text{length} = (r - l) \cdot \min\{height[l], height[r]\} \] +
++ At its core, this is a maximization problem: maximize the contained + area. \[ \max\{(r - l) \cdot \min\{height[l], height[r]\}\} \] +
++ Given a new column position \(l_0 < l\) or \(r_0 < r\), the + contained area can only increase if the height of the corresponding + column increases. +
+ ++ The following correct solution surveys all containers, initialized + with the widest columns positions, that are valid candidates for a + potentially new largest area. A running maximizum, the answer, is + maintained. +
+
+def maxArea(height: list[int]) -> int:
+ area = 0
+ l, r = 0, len(height) - 1
+ while l < r:
+ width, min_height = r - l, min(height[l], height[r])
+ area = max(area, width * min_height)
+ while l < r and height[l] <= min_height:
+ l += 1
+ while l < r and height[r] <= min_height:
+ r -= 1
+ return area
+