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.Iterator;
import java.util.Scanner;
import java.util.HashSet;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
		String S = sc.next();

		HashSet hs = new HashSet();
		for (int i = 0; i < N; i++){
			hs.add(S.charAt(i));
		}

		int cnt = 0;

	    for (int i = 0; i < N; i++) {
	    	for(int j = i + 1; j <= N; j++){
	    		String s1 = S.substring(i, j);
	    		String s2 = S.substring(j);
//	    		System.out.println("j= "+ j +":"+ s1);
//	    		System.out.println("j= "+ j +":"+ s2);
	            Iterator it = hs.iterator();
	            int wkcnt = 0;
	            while (it.hasNext()) {
	            	String chr = it.next().toString();
//	            	System.out.println(chr);
	                if(s1.contains(chr) && s2.contains(chr)){
	                	wkcnt++;
	                }
	            }
	            cnt = Math.max(cnt, wkcnt);
	    	}
	    }


		System.out.println(cnt);

	}
}