# LeetCode 42: Trapping Rain Water — Step-by-Step Visual Trace

**Hard** — Two Pointers | Array | Dynamic Programming | Stack

## The Problem

Given an elevation map represented by an array of non-negative integers, calculate how much water can be trapped after raining.

## Approach

Uses two pointers starting from both ends of the array, tracking the maximum heights seen so far from left and right. Water can be trapped at a position if there are higher bars on both sides, so we calculate trapped water by subtracting current height from the maximum height on the side with the smaller maximum.

**Time:** O(n) · **Space:** O(1)

## Code

```python
class Solution:
    def trap(self, height: List[int]) -> int:
        left, right = 0, len(height) - 1
        max_left, max_right = 0, 0
        trapped_water = 0

        while left < right:
            if height[left] <= height[right]:
                if height[left] >= max_left:
                    max_left = height[left]
                else:
                    trapped_water += max_left - height[left]
                left += 1
            else:
                if height[right] >= max_right:
                    max_right = height[right]
                else:
                    trapped_water += max_right - height[right]
                right -= 1

        return trapped_water
```

## Watch It Run

> **[Open interactive visualization](https://tracelit.dev/app?trace=0042_trapping-rain-water)**

> **Try it yourself:** Open [TraceLit](https://tracelit.dev/app?trace=0042_trapping-rain-water) and step through every line.

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