Add Rain Terraces problem.

Author: trekhleb

src/algorithms/uncategorized/rain-terraces/README.md

@@ -1,6 +1,8 @@
 # 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)
 
@@ -9,44 +11,66 @@ 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**
 
 ```
 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