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.Scanner;
import java.util.Arrays;

public class Main{
  public static void main (String[] args){
    Scanner input = new Scanner(System.in);

    while(input.hasNext()){
      int N = input.nextInt();
      String S = input.next();

      int[] countCharX = new int[26];
      int[] countCharY = new int[26];
      int maxDiff = 0;
      for(int i = 1; i < S.length() - 1; ++i){
        Arrays.fill(countCharX, 0);
        Arrays.fill(countCharY, 0);

        for(int j = 0; j < i; ++j){
          ++countCharX[S.charAt(j) - 'a'];
        }

        for(int j = i; j < S.length(); ++j){
          ++countCharY[S.charAt(j) - 'a'];
        }
        
        int bothContain = 0;
        for(int j = 0; j < countCharX.length; ++j){
          if(countCharX[j] != 0 && countCharY[j] != 0){
            ++bothContain;
          }
        }

        maxDiff = Math.max(maxDiff, bothContain);
     }
     System.out.println(maxDiff);
    } 
  }
}