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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
#include<string.h>

int check(char str2[50], char str3[50]);

int main(void){
	int n;
	scanf("%d", &n);
	
	char str[n];
	scanf("%s", str);
	
	int i;
	int j;
	int c_max;
	int max = 0;
	
	for(i=0;i<=(n-1)/2;i++){
                int k1 = 0;
                int k2 = 0;
		char str2[50] = {'\0'};
		char str3[50] = {'\0'};
		for(j=0;j<n;j++){
			if(j<=i){
				str2[k1++] = str[j];
			}
			else{
				str3[k2++] = str[j];
			}
		}
		c_max = check(str2, str3);
		if(max < c_max){
			max = c_max;
		}
	}
	
	printf("%d", max);
	return 0;
}

int check(char str2[], char str3[]){
	int i;
	int j;
        int k;
	int cnt = 0;
    int flag = 0;
	for(i=0;str2[i]!='\0';i++){
		for(k=0;k<i;k++){
			if(str2[k]==str2[i]){
				flag=1;
				break;
			}
		}
		for(j=0;str3[j]!='\0'&&flag!=1;j++){
			if(str2[i]==str3[j]){
				cnt+=1;
				break;
			}
		}
                flag = 0;
	}
	return cnt;
}