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;

public class Main {
    public static void main(String[] args)throws Exception {
        BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(input.readLine());
        String word = input.readLine();
        int max = -1;
        for (int i = 1; i < n; i++) {
            max = Math.max(max, getDistinct(word.substring(0, i), word.substring( i)));
        }
        System.out.println(max);
    }

    public static int getDistinct(String word1 ,String word2){
        HashSet<Character> wordFreq1 = new HashSet<>();
        for (char c : word1.toCharArray()) {
            wordFreq1.add(c);
        }
        HashSet<Character> wordFreq2 = new HashSet<>();
        for (char c : word2.toCharArray()) {
            wordFreq2.add(c);
        }
        wordFreq1.retainAll(wordFreq2);
        return wordFreq1.size();
    }
}