Problem:
Score : 200 points

Problem Statement

You are given a string S of length N consisting of lowercase English letters.
We will cut this string at one position into two strings X and Y.
Here, we would like to maximize the number of different letters contained in both X and Y.
Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.

Constraints

2 \leq N \leq 100

|S| = N

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the largest possible number of different letters contained in both X and Y.

Sample Input 1

6
aabbca

Sample Output 1

2

If we cut the string between the third and fourth letters into X = aab and Y = bca, the letters contained in both X and Y are a and b.
There will never be three or more different letters contained in both X and Y, so the answer is 2.

Sample Input 2

10
aaaaaaaaaa

Sample Output 2

1

However we divide S, only a will be contained in both X and Y.

Sample Input 3

45
tgxgdqkyjzhyputjjtllptdfxocrylqfqjynmfbfucbir

Sample Output 3

9

Solution in Scala (2.11.7) (judge verdict: Accepted)

import java.util.Scanner

import scala.annotation.tailrec

object Main {
  def main(args: Array[String]): Unit = {
    //nextLineとnextを同時に使ったら死ぬ
    val sc = new Scanner(System.in)

    val n = sc.nextInt()
    val s = sc.next()

    val ans =
      (1 until n)
        .map(i => {
          val a = s.take(i).toCharArray.distinct
          val b = s.drop(i).toCharArray.distinct
          a.count(b.contains(_))
        })
        .max

    println(ans)
  }

  def debug(args: Any*): Unit = {
    Console.err.println(args.mkString(" "))
  }

  def printExecTime(process: => Unit): Unit = {
    val start = System.currentTimeMillis
    process
    debug("処理時間： " + (System.currentTimeMillis - start) + " ミリ秒")
  }

  def bitSearch[T](n: Int, choice: Vector[T]): Vector[Vector[T]] = {
    @tailrec
    def bitSearchLogic(n: Int,
                       choice: Vector[T],
                       acc: Vector[Vector[T]]): Vector[Vector[T]] = {
      if (n == 0) acc
      else {
        val newAcc =
          acc.flatMap(vec => {
            choice.map(c => vec :+ c)
          })
        bitSearchLogic(n - 1, choice, newAcc)
      }
    }

    val firstAcc = choice.map(i => Vector(i))
    bitSearchLogic(n - 1, choice, firstAcc)
  }
}