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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        String S = scan.next();
        System.out.println(solve(N, S));
    }
    public static int solve(int N, String S) {
        char[] data = S.toCharArray();
        int answer = 0;
        for (int i = 0; i < N; i++) {
            answer = Math.max(answer, sol(N, data, i));
        }
        return answer;
    }
    public static int sol(int N, char[] data, int position) {
        HashSet<Character> right = new HashSet<>();
        HashSet<Character> left = new HashSet<>();
        for (int i = 0; i < position; i++) {
            right.add(data[i]);
        }
        for (int i = position; i < N; i++) {
            left.add(data[i]);
        }
        right.retainAll(left);
        return right.size();

    }
}