Step 1 — Setup: Initialize left = 0 and right = n - 1 (two pointers at both ends), and leftMax = 0, rightMax = 0, water = 0. We will always move the pointer that points to the smaller height.

Step 2 — Why move the smaller side: The water level at any index is bounded by the minimum of the maximum height to its left and the maximum height to its right. By moving the pointer at the smaller height, we know the water at that index is bounded by the max we have seen on that side (leftMax or rightMax), so we can safely add water there.

Step 3 — In each iteration: While left < right, compare height[left] and height[right]. If height[left] <= height[right], update leftMax = max(leftMax, height[left]), add max(0, leftMax - height[left]) to water (and optionally record it per index), then left++. Otherwise do the same on the right: update rightMax, add water at right, then right--.

Step 4 — Result: When left meets right, we have considered every index. The total water is the sum of contributions from each index. Time is O(n), space is O(1).