This post discusses a O(n) algorithm that construct a balanced binary search tree (BST) from a sorted list. For instance, consider that we are given a sorted list: [1,2,3,4,5,6,7]. We have to construct a balanced BST as follows.

        4
        |
    2        6 
    |        |
  1   3   5     7

To do so, we use the following definition of Tree, described in Scala By Example book.

abstract class IntSet
case object Empty extends IntSet
case class NonEmpty(elem: Int, left: IntSet, right: IntSet) extends IntSet

One straight-forward approach would be to repeatedly perform binary search on the given list to find the median of the list, and then, to construct a balanced BST recursively. Complexity of such approach is O(nlogn), where n is the number of elements in the given list.

A better algorithm constructs balanced BST while iterating the list only once. It begins with the leaf nodes and construct the tree in a bottom-up manner. As such, it avoids repeated binary searches and achieves better runtime complexity (i.e., O(n), where n is the total number of elements in the given list). Following Scala code outlines this algorithm, which effectively converts a list ls to an IntSet, a balanced BST:

	def toTree(ls: List[Int]): IntSet = {
	def toTreeAux(ls: List[Int], n: Int): (List[Int], IntSet) = {
	if (n <= 0)
	(ls, Empty)
	else {
	val (lls, lt) = toTreeAux(ls, n / 2) // construct left sub-tree
	val x :: xs = lls // extract root node
	val (xr, rt) = toTreeAux(xs, n - n / 2 - 1) // construct right sub-tree

	(xr, IntSet(x, lt, rt)) // construct tree
	}
	}
	val (ls_1, tree) = toTreeAux(ls, List.length(ls))
	tree
	}

view raw toTree.scala hosted with ❤ by GitHub

Any comment or query regarding this post is highly appreciated. Thanks.

This post describes an algorithm to solve UVa 10706: Number Sequence problem from UVa OJ. Before outlining the algorithm, we first give an overview of this problem in the next section.

Interpretation

This is a straight-forward problem, which however, requires careful introspection. Our objective is to write a program that compute the value of a sequence , which is comprised of the number groups . Each consists of positive integers written one after another and thus, the first digits of are —

11212312341234512345612345671234567812345678912345678910123456789101112345678910

It is imperative to note that, the value of has the following range . Realize that, the maximum value of is indeed Int32.MaxValue. In the provided test cases, the values of are stated. We simply have to compute ; therefore, we are required to write a function that has following signature: . That is —

f: 10 --> S.[10] = 4
f: 8  --> S.[8]  = 2
f: 3  --> S.[3]  = 2

Note also that the maximum number of test cases is 25. Question for the reader: What is the maximum possible value of for the specified sequence , which is constituted of ?

We hope that by this time, it is clear where we are headed towards. In next section, we outline an algorithm to solve this problem. But, before moving further, we would like to recommend you to try to solve it yourselves first and then, read the rest of the post. By the way, we believe that the solution that we have presented next can be further optimized (or, even a better solution can be derived). We highly welcome your feedback regarding this.

Algorithm

Recall that, is constituted of number groups . In order to identify the digit located at the position of , we first determine the number group that digit is associated with and then, we figure out the digit from .

To do so, we first compute the length of each number group . Consider the number group till —

112123123412345123456…12345678910

As mentioned, this sequence is basically constituted of as shown below (with their respective lengths).

1                    --> 1 
1 2                  --> 2 
1 2 3                --> 3
1 2 3 4              --> 4
1 2 3 4 5            --> 5
1 2 3 4 5 6          --> 6
1 2 3 4 5 6 7        --> 7
1 2 3 4 5 6 7 8      --> 8
1 2 3 4 5 6 7 8 9    --> 9
1 2 3 4 5 6 7 8 9 10 --> 11

It implies that the length of can be computed from the length of as shown below.

Using Eq.1, we calculate the cumulative sum of each number group as follows.

Why do we calculate cumulative sum? The purpose in this is to be able to simply run a binary search to determine which number group the digit belongs to in time. For example, consider i=40.

1                    --> 1 
1 2                  --> 2 
1 2 3                --> 6 
1 2 3 4              --> 10
1 2 3 4 5            --> 15
1 2 3 4 5 6          --> 21
1 2 3 4 5 6 7        --> 28
1 2 3 4 5 6 7 8      --> 36 
1 2 3 4 5 6 7 8 9    --> 45
1 2 3 4 5 6 7 8 9 10 --> 56

Using binary search, we can find out that contains the digit. Then, using a linear search, we can simply derive the first four digits of to eventually find out the corresponding digit, 4.

Similarly, for i=55, we can figure out that the digit is indeed 1.

Implementation

Next, we outline a F# implementation of the stated algorithm. We start with computing the cumulative sums of the lengths as follows. Continue reading “UVa 10706. Number Sequence with F#”