# LeetCode 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 is able to trap after raining.

For example,
Given `[0,1,0,2,1,0,1,3,2,1,2,1]`, return `6`. 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!

## Explanation

The key point is to find the relationship between bar heights and amount of the water.

We can find the relationship by looking at each individual bar and get the amount of water it can hold. Then sum up all water bar hold together.

## Java Solution

```class Solution {
public int trap(int[] height) {
int totalAmount = 0;
if (height == null || height.length == 0) {
}

int[] leftHighest = new int[height.length + 1];
leftHighest = 0;
for (int i = 0; i < height.length; i++) {
leftHighest[i + 1] = Math.max(leftHighest[i], height[i]);
}

int rightHighest = 0;
for (int i = height.length - 1; i >= 0; i--) {
rightHighest = Math.max(height[i], rightHighest);
totalAmount += Math.min(leftHighest[i], rightHighest) > height[i] ? Math.min(leftHighest[i], rightHighest) - height[i] : 0;
}

1. mriganka bora says: