> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/strings-arrays-and-2-pointers/trapping-rain-water.md).

# Trapping Rain Water

Given *n* non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

\
The above elevation map 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. **Thanks Marcos** for contributing this image!

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

**Example:**

```
Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
```

```java
class Solution {
    // O(N) Space
    public int trap(int[] height) {
        int maxLeft[] = new int[height.length];
        for (int i = 0; i < height.length; i++) {
            if (i == 0)
                maxLeft[i] = height[i];
            else
                maxLeft[i] = Math.max(height[i], maxLeft[i - 1]);
        }

        int[] maxRight = new int[height.length];
        for (int i = height.length - 1; i >= 0; i--) {
            if (i == height.length - 1)
                maxRight[i] = height[i];
            else
                maxRight[i] = Math.max(height[i], maxRight[i + 1]);
        }

        int sum = 0;
        for (int i = 0; i < height.length; i++) {
            // Now for every wall, we are finding the max wall on it's right & left (including itself)
            // Now the water above this wall can be the min(maxleft,maxright)
            int minWall = Math.min(maxLeft[i], maxRight[i]);
            sum += minWall - height[i];
        }
        return sum;
    }
    // O(1) Space
    // https://drive.google.com/file/d/1wvPyepc1cw2IPgtK9TmjjBzSIB6uVy8t/view?usp=sharing
}
```