Hackerrank – Problem Statement
A description of the problem can be found on Hackerrank.
Solution
I started with finding all unique characters.
Then I created all possible combinations of pairs.
- I deleted other characters except the pair.
- I found if the examined pair contains alternating characters in deleted strings
- If yes I count the length of the string.
The result is the maximum of string lengths.
I created solution in:
All solutions are also available on my GitHub profile.
Scala
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 |
import scala.io.Source object Solution extends App { val lines = Source.stdin.getLines().toList val s = lines.tail.head val characters = s.distinct val strings = (for { i <- 0 until characters.length - 1 j <- i + 1 until characters.length } yield { val c1 = characters(i) val c2 = characters(j) val subset = s.replaceAll("[^" + c1 + "" + c2 + "]", "") subset }).filter(isAlternating) val maxLength = if(strings.isEmpty) 0 else strings.maxBy(_.length).length println(maxLength) def isAlternating(s: String): Boolean = s.sliding(2).forall(ss => ss(0) != ss(1)) } |
Java
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 |
import java.util.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int len = in.nextInt(); String s = in.next(); Set<Character> distinct = new HashSet<>(); for (int i = 0; i < len; i++) { distinct.add(s.charAt(i)); } List<Character> distinctList = new ArrayList<>(distinct); int max = 0; for(int i = 0; i < distinct.size() - 1; i++) { for(int j = i + 1; j < distinct.size() ; j++) { char c1 = distinctList.get(i); char c2 = distinctList.get(j); String subset = s.replaceAll("[^" + c1 + "" + c2 + "]", ""); if(isAlternating(subset)) { int l = subset.length(); max = l > max ? l : max; } } } System.out.println(max); } private static boolean isAlternating(String s) { for(int i = 0; i < s.length() - 1; i++) { if(s.charAt(i) == s.charAt(i + 1)) { return false; } } return true; } } |
Javascript
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 |
process.stdin.resume(); process.stdin.setEncoding('ascii'); var input_stdin = ""; var input_stdin_array = ""; var input_currentline = 0; process.stdin.on('data', function (data) { input_stdin += data; }); process.stdin.on('end', function () { input_stdin_array = input_stdin.split("\n"); main(); }); function readLine() { return input_stdin_array[input_currentline++]; } /////////////// ignore above this line //////////////////// function isAlternating(s) { for(let i = 0; i < s.length - 1; i++) { if(s.charAt(i) == s.charAt(i + 1)) { return false; } } return true; } function maxLen(n, s){ var distinct = {}; for (var i = 0; i < n; i++) { distinct[s[i]] = 1; } distinct = Object.keys(distinct); var max = 0; for(var i = 0; i < distinct.length - 1; i++) { for(var j = i + 1; j < distinct.length ; j++) { var c1 = distinct[i]; var c2 = distinct[j]; var subset = s.replace(new RegExp('[^' + c1 + '' + c2 + ']', "gi"), ''); if(isAlternating(subset)) { var l = subset.length; max = l > max ? l : max; } } } return max; } function main() { var n = parseInt(readLine()); var s = readLine(); var result = maxLen(n, s); process.stdout.write(""+result+"\n"); } |