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.*;

class Main{
	int n;
  String s;
  

  public static void main(String[] ar){
    new Main();
  }
  public Main(){
  	Scanner sc=new Scanner(System.in);
  	n=Integer.parseInt(sc.nextLine());
    s=sc.nextLine();

    int result=0;
    for(int i=1;i<n;i++){
      //System.out.println(s.substring(0,i)+" "+s.substring(i));
      int count=countChar(s.substring(0,i),s.substring(i));
      if(result<count) result=count;
    }
    System.out.println(result);

    //System.out.println(countChar(s));
  }
  public int countChar(String str1,String str2){
    int c1[]=new int[26];
    int c2[]=new int[26];

    for(int i=0;i<str1.length();i++){
      c1[(int)str1.charAt(i)-97]=1;
    }

    for(int i=0;i<str2.length();i++){
      c2[(int)str2.charAt(i)-97]=1;
    }

    int sum=0;
    for(int i=0;i<26;i++){
      //System.out.println(c1[i]+" "+c2[i]);
      sum+=c1[i]*c2[i];
    }
    return sum;
  }
}