Decode the Coding Interview in Scala: Real-World Examples/

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Feature #11: Generate Movie Viewing Orders

Feature #11: Generate Movie Viewing Orders

Implementing the "Generate Movie Viewing Orders" feature for our "Netflix" project.

We'll cover the following...

Description

We want to offer marathons for our viewers. Each marathon will have a fixed set of movies catering to a specific taste. For a given marathon, different viewing orders of the movies will yield different user satisfaction results. We want to experiment (A/B testing) with different viewing orders of the same marathon.

Your task is to generate all the possible permutations of movies in a given marathon.

Let’s look at an example to better understand this:

Solution

To solve this problem, we will use the backtracking approach.

We will assume a backtrack function that takes the index of the first movie to consider as an argument backtrack(first).

  • If the first movie to consider has index n, then that means that the current permutation is done.

  • We will iterate over the marathon from index first to index n - 1.

  • We will place the ith movie first in the permutation, that is, movies[first], movies[i] = movies[i], movies[first].

  • We will proceed to create all the permutations that start from the ith movie: backtrack(First + 1).

  • Now we will backtrack, that is, movies[First], movies[i] = movies[i], movies[First] back.

Let’s look at some illustrations to better understand this:

Note: We are using numbers to represent movies in the following illustration.

Let’s look at the code for this solution:

import scala.collection.mutable._
object Solution {
  def swap(first:Int, i:Int, moviesList:ArrayBuffer[String]):Unit={
    var temp = moviesList(first)
    var temp2 = moviesList(i)
    moviesList.update(i, temp)
    moviesList.update(first, temp2)
  }
  def backTrack(first: Int, size: Int, moviesList: ArrayBuffer[String], output: ArrayBuffer[ArrayBuffer[String]]): Unit = { // If all strings of given array `moviesList` are used and
    // and Backtracking is performed add the permutations to output array.
    if (first == size) {
      val temp = new ArrayBuffer[String] ++ moviesList
      output.addOne(temp)
    }
    // Perform Backtracking for the size of a given array.
    for (i <- first until size) { // Swap: In the current permutation place i-th integer first.
      swap(first, i, moviesList)
      // Complete permutations using the next integers.
      backTrack(first + 1, size, moviesList, output)
      // Swap and Backtrack
      swap(first, i, moviesList)
    }
  }
  def generatePermutations(movies: Array[String]): ArrayBuffer[ArrayBuffer[String]] = {
    val output = new ArrayBuffer[ArrayBuffer[String]]
    val size = movies.length
    // convert movies into list since the output is a list of lists
    val moviesList = new ArrayBuffer[String]
    for (movie <- movies) {
      moviesList.addOne(movie)
    }
    backTrack(0, size, moviesList, output)
    output
  }
  def printArrayBuffer(list:ArrayBuffer[ArrayBuffer[String]]):Unit={
    print("[")
    for(l <- list){
      print("[" + l.mkString(", ") + "]")
    }
    println("]")
  }
  def main(args: Array[String]): Unit = {
    // Example #1
    val input = Array("Frozen", "Dune", "Coco")
    var output = generatePermutations(input)
    println("Output 1: ")
    printArrayBuffer(output)
    // Example #2
    val input2 = Array("Frozen", "Dune", "Coco", "Melificient")
    output = generatePermutations(input2)
    println("Output 2: ")
    printArrayBuffer(output)
    // Example #3
    val input3 = Array("Dune", "Coco")
    output = generatePermutations(input3)
    println("Output 3: ")
    printArrayBuffer(output)
  }
}
Permutations

Complexity measures

Time complexity
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