Password Cracker FP – HackerRank Solution

In this post, we will solve Password Cracker FP HackerRank Solution. This problem (Password Cracker FP) is a part of HackerRank Functional Programming series.


There are N users registered on a website Each of them have a unique password represented by pass[1], pass[2], …, pass[N]. As this a very lovely site, many people want to access those awesomely cute pics of the kittens. But the adamant admin don’t want this site to be available for general public. So only those people with passwords can access it.

Yu being an awesome hacker finds a loophole in their password verification system. A string which is concatenation of one or more passwords, in any order, is also accepted by the password verification system. Any password can appear 0 or more times in that string. He has access to each of the N passwords, and also have a string loginAttempt, he has to tell whether this string be accepted by the password verification system of the website.

For example, if there are 3 users with password {"abra""ka""dabra"}, then some of the valid combinations are "abra" (pass[1])"kaabra" (pass[2]+pass[1])"kadabraka" (pass[2]+pass[3]+pass[2])"kadabraabra" (pass[2]+pass[3]+pass[1]) and so on.

Input Format

First line contains an integer T, the total number of test cases. Then T test cases follow.
First line of each test case contains N, the number of users with passwords. Second line contains N space separated strings, pass[1] pass[2] … pass[N], representing the passwords of each user. Third line contains a string, loginAttempt, for which Yu has to tell whether it will be accepted or not.


  • 1 <= T <= 10
  • 1 <= N <= 10
  • pass[i] != pass[j], 1 <= i < j <= N
  • 1 <= length(pass[i]) <= 10, where i = [1, N]
  • 1 < length(loginAttempt) <= 2000
  • loginAttempt and pass[i] contains only lowercase latin characters (‘a’-‘z’).

Output Format

For each valid string, Yu has to print the actual order of passwords, separated by space, whose concatenation results into loginAttempt. If there are multple solutions, print any of them. If loginAttempt can’t be accepted by the password verification system, then print WRONG PASSWORD.

Sample Input 0

because can do must we what
hello planet
ab abcd cd

Sample Output 0

we do what we must because we can
ab cd

Explanation 0

Sample Case #00: "wedowhatwemustbecausewecan" is the concatenation of passwords {"we""do""what""we""must""because""we""can"}. That is

loginAttempt = pass[5] + pass[3] + pass[6] + pass[5] +  pass[4] + pass[1] + pass[5] + pass[2]

Note that any password can repeat any number of times.

Sample Case #01: We can’t create string "helloworld" using the strings {"hello""planet"}.

Sample Case #02: There are two ways to create loginAttempt ("abcd"). Both pass[2] = "abcd" and pass[1] + pass[3] = "ab cd" are valid answers.

Sample Input 1

ozkxyhkcst xvglh hpdnb zfzahm
gurwgrb maqz holpkhqx aowypvopu
a aa aaa aaaa aaaaa aaaaaa aaaaaaa aaaaaaaa aaaaaaaaa aaaaaaaaaa

Sample Output 1


Solution – Password Cracker FP – HackerRank Solution


import java.util.Scanner

import scala.collection.mutable

object Solution {
  private val data = mutable.Map[String, List[String]]()

  def main(args: Array[String]): Unit = {
    val sc = new Scanner(

    val t = sc.nextInt
    (0 until t).foreach(_ => {
      val passwords = sc.nextLine.split(" ")
      val loginAttempt = sc.nextLine

      println(solve(loginAttempt, passwords.toIndexedSeq))


  def solve(loginAttempt: String, passwords: Seq[String]): String = {

    def inner(loginAttempt: String, acc: List[String]): List[String] = data.getOrElseUpdate(loginAttempt,
      if (loginAttempt.isEmpty) acc
      else {
        passwords.foldLeft(List[String]())((list, password) => if (list.isEmpty) {
          val isPrefix = loginAttempt.startsWith(password)

          val rest = if (isPrefix) inner(loginAttempt.substring(password.length), password :: acc) else Nil

          if (isPrefix && rest.nonEmpty) rest else Nil
        } else list)

    val answer = inner(loginAttempt, Nil).reverse
    if (answer.isEmpty) "WRONG PASSWORD" else answer.mkString(" ")

Note: This problem (Password Cracker FP) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.

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