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.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

import javax.print.event.PrintJobListener;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.next();
        int result = 0;
        Set<String> set = new HashSet<>();
        for (int i = 1; i < n; i++) {
            List<String> ss = new ArrayList<>();
            char[] a = s.substring(i).toCharArray();
            char[] b = s.substring(0, i).toCharArray();
            int tmp = 0;
            for (char c : b) {
                if (String.valueOf(a).contains(String.valueOf(c)) && !ss.contains(String.valueOf(c))) {
                    tmp++;
                    ss.add(String.valueOf(c));
                }
            }
            result = Math.max(result, tmp);
        }
        System.out.println(result);
    }
}