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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {

	public static void main(String[] args) throws IOException {
		
		// 入出力高速化 java.io.* とthrows IOExceptionを忘れずに
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int N = Integer.parseInt(br.readLine());
		String[] IN =br.readLine().split("") ; 
		
		int MAX = 0;
		
		for(int i=0; i<N; i++) {
			
			HashMap<String,Integer> count = new HashMap<String,Integer>();
			//文字列をカットした左半分について
			for(int j=0; j<i+1; j++) {
				if (!count.containsKey(IN[j])) {
					count.put(IN[j], 1); //まだHashMapにキーが登録されていないなら、登録する
				}
			}
			
			//文字列をカットした右半分について
			for(int j=i+1; j<N; j++) {
				if (count.containsKey(IN[j])) { //HashMapに既にキーがあるとき
					int now = count.get(IN[j]); //キーSに紐づく中身のintを取得
					count.replace(IN[j], now + 1);
				}
			}
			
			int c = 0;
			for(int key1:count.values()) {
				if(key1 > 1) {
					c++;
				}
			}
			if(MAX<c) {
				MAX = c;
			}
		}
		
		System.out.println(MAX);	
	}
}