Trapping rain water with two pointers

You're given n non-negative bar heights. After rain, water pools in the dips between taller bars. The water sitting above bar i is bounded by the tallest bar to its left and the tallest bar to its right: it equals $\min(\text{maxLeft}_i, \text{maxRight}_i) - \text{height}_i$ (never negative).

height = [3, 0, 2, 0, 4]
-> 7           # 3 + 1 + 3 units above the three dips

Return the total trapped water. Aim for $O(n)$ time and $O(1)$ space, and justify why the pointer you advance is the safe one.