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.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String s = sc.next();
		Set<String> first = new HashSet<>();
		Set<String> second = new HashSet<>();

		int max = 0;
		for (int i = 1; i < n; i++) {
			first.clear();
			second.clear();

			String firstStr = s.substring(0, i);
			String secondStr = s.substring(i, n);

			for (char c : firstStr.toCharArray()) {
				first.add(String.valueOf(c));
			}
			for (char c : secondStr.toCharArray()) {
				second.add(String.valueOf(c));
			}

			int cnt = 0;
			for (Iterator<String> iterator = first.iterator(); iterator.hasNext();) {
				String str = iterator.next();
				if (second.contains(str)) cnt++;
			}

			max = Math.max(max, cnt);

		}

		System.out.println(max);

		sc.close();

	}

}