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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    String S = sc.next();

    int countMax = 0 ;

    for(int i = 1;i < S.length()  ; i ++){
      int count = 0 ;
      int[] frontChar = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
      int[] backChar  = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
      String front = S.substring(0,i );
      String back  = S.substring(i  );
      for(int j = 0 ;j < front.length(); j++){
        if(frontChar[ (int) front.charAt(j)  - 97 ] == 0 ) {
          frontChar[ (int) front.charAt(j)  - 97 ] ++ ;
        }
      }
      for(int j = 0 ;j < back.length(); j++){
        if(backChar[ (int) back.charAt(j)  - 97 ] == 0 ) {
          backChar[ (int) back.charAt(j)  - 97 ] ++ ;
        }
      }
      for(int k = 0 ; k < 26 ; k ++){
        if(frontChar[k]  == 1 && backChar[k] == 1){
          count ++ ;
        }
      }
      if(countMax < count ){
        countMax = count ;
      }
    }

    System.out.println( countMax );
  }
}