Intuition

Water sitting above a bar equals min(tallest wall on its left, tallest wall on its right) - the bar’s own height.

With two pointers we always process the side whose max is definitely the limiting one: if the left bar is shorter than the right bar, the left side’s water depends only on leftMax, so we can settle it immediately.

Approach

Pointers l, r plus leftMax, rightMax.
If height[l] < height[r]: update leftMax, add leftMax - height[l], move l right.
Else: update rightMax, add rightMax - height[r], move r left.
Accumulate until the pointers meet.

Walkthrough

height = [4, 2, 0, 3, 2, 5]
right side lower until the 5 at the end;
trapped: above the 2 -> 2, above 0 -> 3, above 3 -> 0, above 2 -> 1... total = 9

Solution

def trap(height):
    l, r = 0, len(height) - 1
    leftMax = rightMax = water = 0
    while l < r:
        if height[l] < height[r]:
            leftMax = max(leftMax, height[l])
            water += leftMax - height[l]
            l += 1
        else:
            rightMax = max(rightMax, height[r])
            water += rightMax - height[r]
            r -= 1
    return water

Complexity

Time — O(n): a single two-pointer pass.
Space — O(1): a few scalars (the precompute-arrays version costs O(n)).

Common Pitfalls

Update the running max before adding water, or you may subtract more than the wall height.

Trapping Rain Water

Example 1

Example 2

Constraints