# 0042. Trapping Rain Water

<https://leetcode.com/problems/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.

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png)

```
**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 * 104`
* `0 <= height[i] <= 105`

## ac1: two pointers

[https://leetcode.com/problems/trapping-rain-water/discuss/17357/Sharing-my-simple-c%2B%2B-code%3A-O(n)-time-O(1)-space](https://leetcode.com/problems/trapping-rain-water/discuss/17357/Sharing-my-simple-c%2B%2B-code%3A-O%28n%29-time-O%281%29-space)

```java
class Solution {
    public int trap(int[] height) {
        int res = 0;
        int leftMax = 0, rightMax = 0, left = 0, right = height.length - 1;
        
        while (left <= right) {
            if (leftMax < rightMax) {
                if (height[left] >= leftMax) {
                    leftMax = height[left];
                } else {
                    res += leftMax - height[left];
                }
                left++;
            } else {
                if (height[right] >= rightMax) {
                    rightMax = height[right];
                } else {
                    res += rightMax - height[right];
                } 
                right--;
            }
        }
        
        return res;
    }
}
```

## ac2: brute force

```java
class Solution {
    public int trap(int[] height) {
        int n = height.length;
        // edge cases
        if (n < 3) return 0;

        int water = 0;
        // 1st pass walk through, 2nd pass find left max, 3rd pass find right max
        for (int i = 0; i < n; i++) {
            // find left max
            int lmax = 0;
            for (int l = 0; l <= i; l++) {
                lmax = Math.max(lmax, height[l]);
            }
            // find right max
            int rmax = 0;
            for (int r = n-1; r >= i; r--) {
                rmax = Math.max(rmax, height[r]);
            }
            // calculate water volumn
            water += Math.min(rmax, lmax) - height[i];
        }

        // result
        return water;
    }
}

/*
brute force, O(n2) time + O(1) space
DP-like memorization + 3 pass, O(3n) time + O(2n) space
stack + 1 pass, O(n) time + O(n) space
two pointers + 1 pass, O(n) time + O(1) space
*/
```

## ac3: DP-like memorization

```java
class Solution {
    public int trap(int[] height) {
        int n = height.length;
        // edge cases
        if (n < 3) return 0;

        int[] lmax = new int[n];
        int[] rmax = new int[n];
        lmax[0] = height[0];
        rmax[n-1] = height[n-1];

        // find left max
        for (int i = 1; i < n; i++) {
            lmax[i] = Math.max(lmax[i-1], height[i]);
        }
        // find right max
        for (int i = n - 2; i >= 0; i--) {
            rmax[i] = Math.max(rmax[i+1], height[i]);
        }

        // accumulate water
        int water = 0;
        for (int i = 0; i < n; i++) {
            water += Math.min(lmax[i], rmax[i]) - height[i];
        }

        // result
        return water;
    }
}
```

## ac4: stack

Very hard to understand, doubt if it's suitable for interview.\
<https://leetcode.com/problems/trapping-rain-water/discuss/17414/A-stack-based-solution-for-reference-inspired-by-Histogram>

```java
class Solution {
    public int trap(int[] height) {
        int n = height.length;
        int water = 0;
        // edge cases
        if (n < 3) return 0;

        // stack 
        Stack<Integer> stack = new Stack<Integer>();
        for (int i = 0; i < n; i++) {
            while (!stack.isEmpty() && height[i] > height[stack.peek()]) {
                int bottom = height[stack.pop()];
                if (stack.isEmpty()) break;
                int length = i - stack.peek() - 1;
                int h = Math.min(height[i], height[stack.peek()]) - bottom;
                water += length * h;
            }
            stack.push(i);
        }

        // result
        return water;
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://jaywin.gitbook.io/leetcode/solutions/0042-trapping-rain-water.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.