leetcode/apps/docs/content/(01-linear-structures)/two-pointers/(problems)/42_trapping-rain-water.mdx

---
title: '42. Trapping Rain Water'
description: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Two Pointers", slug: "two-pointers" }
  - { name: "Dynamic Programming", slug: "dynamic-programming" }
  - { name: "Stack", slug: "stack" }
  - { name: "Monotonic Stack", slug: "monotonic-stack" }
issue: "https://github.com/prdlk/leetcode/issues/211"
sidebar:
  label: 'Trapping Rain Water'
  badge: 'Hard'
---

### Example 1:
- Input: `height = [0,1,0,2,1,0,1,3,2,1,2,1]`
- Output: `6`
- Explanation: The above elevation map (black section) is represented by array `[0,1,0,2,1,0,1,3,2,1,2,1]`. In this case, `6` units of rain water (blue section) are being trapped.

### Example 2:
- Input: `height = [4,2,0,3,2,5]`
- Output: `9`

### Constraints:

- `n == height.length`
- `1 <= n <= 2 * 10^4`
- `0 <= height[i] <= 10^5`

## Solution

```py
class Solution:
    def trap(self, height: list[int]) -> int:
        left, right = 0, len(height) - 1
        leftMax, rightMax = 0, 0
        total = 0

        while left < right:
            leftMax = max(leftMax, height[left])
            rightMax = max(rightMax, height[right])

            if height[left] < height[right]:
                total += leftMax - height[left]
                left += 1
            else:
                total += rightMax - height[right]
                right -= 1

        return total
```