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;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();//, n = sc.nextInt();
		int sum=0,count=0,flag=0,max=-999999,min=99999;
		//String[] array = new String[n];
		String s=sc.next();//,t=sc.next();
		String buff1,buff2;

		for(int i=0;i<s.length()-1;i++) {
			buff1=s.substring(0,i+1);
			buff2=s.substring(i+1,s.length());
			//System.out.println(buff1+" "+buff2);
			for(int j=0;j<buff1.length();j++) {
				for(int l=0;l<j;l++) {
					if(buff1.charAt(j)==buff1.charAt(l)) {
						flag=1;
						//System.out.println("same");
					}
				}
				if(flag==1) {
					flag=0;
					continue;
				}
				flag=0;
				for(int k=0;k<buff2.length();k++) {
					if(buff1.charAt(j)==buff2.charAt(k)) {
						count++;
						break;
					}
				}
			}
			//System.out.println(count);
			max=Math.max(count, max);
			count=0;
		}

		System.out.println(max);

	}
	public static int sample(int n) {
		return n;
	}

}