Question

Scala: Write and test a Program as elegantly as possible

Given a piece of text, create a histogram of letter pairs (order from high to low). For instance, for the text, “this is a good thing”,

the letter pairs are: th, hi, is, is, go, oo, od, th, hi, in, and ng. (ignore a) The histogram will be:
th: 2, is: 2, hi: 2 go: 1, oo: 1, od: 1, in: 1, ng: 1

Sample Input/Output:

Enter text: this is a good thing
Histogram: th: 2, is: 2, hi: 2 go: 1, oo: 1, od: 1, in: 1, ng: 1

Enter text: coooooool
Histogram: oo: 6, co: 1, ol: 1

Answer 1

Code in Scala:

import scala.collection.immutable.ListMap //immutable listmap cannot be updated
import scala.collection.mutable.Map //mutable map can be updated

object Histogram {

  def main(args: Array[String]): Unit =
  {
    println("Enter text: ") //Print message
    val string = scala.io.StdIn.readLine().toLowerCase //Take input from user and process it to lower case
    var word_list: List[String] = string.split("\\s+").toList //Split given string by white spaces e.g. To do => ['To', 'do']
    word_list = word_list.filter(_.length > 1) //Eliminate words with character length = 1
    val map:Map[String, Int] = Map() //Create a map with key as String and value as Integer
    for(word <- word_list) //Iterate over list of words
      {
        for(character_index <- 1 until word.length) //Iterate over each word
          {
            val key = word.charAt(character_index - 1).toString + word.charAt(character_index) //Take two characters from string and concatenate them
            map(key) = if(map.contains(key)) map(key) + 1 else 1 //If key exists, increment current count otherwise create entry
          }
      }

    val sorted_map = ListMap(map.toSeq.sortWith(_._2 > _._2):_*) //Sort map according to values in descending order
    print("Histogram:\t") //Display message
    for((key, value) <- sorted_map) //Iterate over sorted map
    {
      print(key+":"+value+"\t") //Print key:value pair
    }
  }
}

Code Description:
Input: String
Output: List of substrings with frequency count

Main Logic:

Input: This is a good thing

Execution:

1. Take input string from user (This is a good thing)
2. Convert the string to lowercase (this is a good thing)
3. Split the string according to one or more (\\s+) white spaces (['this', 'is', 'a', 'good', 'thing'])
4. Remove strings with length 1 (['this', 'is', 'good', 'thing'])
5. Initialize a map to store substrings and their counts
6. Iterate over list created from Step 4 (['this', 'is', 'good', 'thing'])
   1. ​​​​​​​Iterate over each character of current string (['t', 'h', 'i', 's'])
      • Concatenate neighboring characters and create substrings (['th', 'hi', 'is'])
      • Check whether substring is present in map otherwise create an entry
        • If substring is present, increment the existing count by 1
7. Sort items of map according to values from (key, value) pair and follow descending order
8. Iterate over sorted map and print substring and frequency count