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we need to fetch and display each new word after each conversion until we reach our end result. We’ll traverse the array and return the complete sequence of words after each conversion from start to end.

Sumit Rawal answered on September 8, 2023 Popularity 1/10 Helpfulness 1/10

answer we need to fetch and display each new word after each conversion until we reach our end result. We’ll traverse the array and return the complete sequence of words after each conversion from start to end.

we need to fetch and display each new word after each conversion until we reach our end result. We’ll traverse the array and return the complete sequence of words after each conversion from start to end.

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import scala.collection.mutable.Queue
import scala.collection.mutable.Map

object WordConversionWithPath {
  def main(args: Array[String]): Unit = {
    val initialWord = "hit"
    val finalWord = "cog"
    val wordList = Array("hot", "dot", "dog", "lot", "log", "cog")

    val transformationSequence = findTransformationSequence(initialWord, finalWord, wordList)

    if (transformationSequence.isEmpty)
      println(s"No transformation sequence exists from $initialWord to $finalWord.")
    else {
      println(s"Transformation sequence from $initialWord to $finalWord:")
      transformationSequence.foreach(println)
    }
  }

  def findTransformationSequence(initialWord: String, finalWord: String, wordList: Array[String]): List[String] = {
    if (!wordList.contains(finalWord))
      return List.empty // The final word is not in the word list; transformation is not possible.

    val graph = Map[String, List[String]]()

    // Build the graph by creating edges between words that differ by one character.
    for (word1 <- wordList) {
      graph(word1) = wordList.filter(word2 => isOneCharApart(word1, word2)).toList
    }

    val visited = Map[String, Boolean]()
    visited(initialWord) = true

    val queue = Queue[(String, List[String])]()
    queue.enqueue((initialWord, List(initialWord))) // Initial word is at level 1.

    while (queue.nonEmpty) {
      val (currentWord, pathSoFar) = queue.dequeue()

      for (neighbor <- graph(currentWord)) {
        if (neighbor == finalWord)
          return pathSoFar :+ finalWord // Reached the final word.

        if (!visited.contains(neighbor)) {
          visited(neighbor) = true
          queue.enqueue((neighbor, pathSoFar :+ neighbor))
        }
      }
    }

    List.empty // No transformation sequence exists.
  }

  def isOneCharApart(word1: String, word2: String): Boolean = {
    var diffCount = 0

    for (i <- 0 until word1.length) {
      if (word1(i) != word2(i)) {
        diffCount += 1
        if (diffCount > 1) {
          return false
        }
      }
    }

    diffCount == 1
  }
}

Popularity 1/10 Helpfulness 1/10 Language scala

Source: chat.openai.com

Tags: display each fetch return scala sequence word words

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Contributed on Sep 08 2023

Sumit Rawal

0 Answers Avg Quality 2/10

we need to fetch and display each new word after each conversion until we reach our end result. We’ll traverse the array and return the complete sequence of words after each conversion from start to end.

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we need to fetch and display each new word after each conversion until we reach our end result. We’ll traverse the array and return the complete sequence of words after each conversion from start to end.

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