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.HashSet;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        sc.nextLine();
        String s = sc.nextLine();
        int[] ans= new int[n-1];
        HashSet<Character> st = new HashSet<Character>();

        sc.close();
        for(int i=1; i<s.length()-1; i++) {
            st.clear();
            ans[i-1] = 0;
            for(int j=0; j<i; j++) {
                for(int k=i; k<s.length(); k++) {
                    if(!(st.contains(s.charAt(j))) && s.charAt(j)==s.charAt(k)) {
                        ans[i-1]++;
                        st.add(s.charAt(j));
                        break;
                    }
                }
            }
        }
        int max = 0;
        for(int i: ans) {
            if(i>max) max=i;
        }
        System.out.println(max);
    }
}