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Add Rain Terraces problem.

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Oleksii Trekhleb 2018-07-27 12:59:56 +03:00
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src/algorithms/uncategorized/rain-terraces/README.md
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# Rain Terraces (Trapping Rain Water) Problem
Given an array of non-negative integers representing terraces in an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
Given an array of non-negative integers representing terraces in an elevation map
where the width of each bar is `1`, compute how much water it is able to trap
after raining.
![Rain Terraces](https://www.geeksforgeeks.org/wp-content/uploads/watertrap.png)
@ -9,27 +11,31 @@ Given an array of non-negative integers representing terraces in an elevation ma
**Example #1**
```
Input: arr[] = [2, 0, 2]
Input: arr[] = [2, 0, 2]
Output: 2
Structure is like below
Structure is like below:
| |
|_|
We can trap 2 units of water in the middle gap.
```
**Example #2**
```
Input: arr[] = [3, 0, 0, 2, 0, 4]
Input: arr[] = [3, 0, 0, 2, 0, 4]
Output: 10
Structure is like below
Structure is like below:
|
| |
| | |
|__|_|
We can trap "3*2 units" of water between 3 an 2,
"1 unit" on top of bar 2 and "3 units" between 2
and 4. See below diagram also.
and 4. See below diagram also.
```
**Example #3**
@ -37,16 +43,34 @@ and 4. See below diagram also.
```
Input: arr[] = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
Output: 6
Structure is like below:
|
| || |
_|_||_||||||
Trap "1 unit" between first 1 and 2, "4 units" between
first 2 and 3 and "1 unit" between second last 1 and last 2
first 2 and 3 and "1 unit" between second last 1 and last 2.
```
## Algorithms
## The Algorithm
An element of array can store water if there are higher bars on left and right.
We can find amount of water to be stored in every element by finding the heights
of bars on left and right sides. The idea is to compute amount of water that can
be stored in every element of array. For example, consider the array
`[3, 0, 0, 2, 0, 4]`, We can trap "3*2 units" of water between 3 an 2, "1 unit"
on top of bar 2 and "3 units" between 2 and 4. See below diagram also.
A **simple solution** is to traverse every array element and find the highest
bars on left and right sides. Take the smaller of two heights. The difference
between smaller height and height of current element is the amount of water
that can be stored in this array element. Time complexity of this solution
is `O(n2)`.
An **efficient solution** is to pre-compute highest bar on left and right of
every bar in `O(n)` time. Then use these pre-computed values to find the
amount of water in every array element.
## References