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: Accepted)

#include <stdio.h>

int main(int argc, const char * argv[]) {
    
    int n = 0;//長さ　6
    scanf("%d",&n);
    
    char s[n];//文字列
    scanf("%s",s);
    
    char strA[100];
    char strB[100];
    char hikaku[100];
    int result = 0;
    int count = 0;
    
    
    
    for (int i = 0; i < n-1; i++) {
        strA[i] = s[i];
//        printf("%c,",s[i]);
        for (int j = i+1; j < n; j++) {//0  1,2,3,4,5
            strB[j-(i+1)] = s[j];
//            printf("%c",s[j]);
        }
//        printf(" ");
        //1:5,2:4,3:3,4:2,5:1ができてるはず
        
        for(int ii = 0; ii < i+1; ii++){
            for(int jj = 0; jj < n-(i+1); jj++){
//                printf("strB = %c  ",strB[jj]);
                if (strA[ii] == strB[jj]) {
//                    printf("strA = %c,strB = %c",strA[ii],strB[jj]);
                    //同じのをカウントしない
                    if (count == 0) {
                        hikaku[count] = strA[ii];
                        count++;
                        break;
                    }else{
                        int sameFlag = 0;
                        for (int k = 0; k < count; k++) {
                            if (hikaku[k] == strA[ii]) {
                                sameFlag = 1;
                                break;
                            }
                        }
                        if (sameFlag == 1) {
                            break;
                        }else{
                            hikaku[count] = strA[ii];
                            count++;
                            break;
                        }
                    }
                }
            }
        }
        
        
        if(count >= result){
            result = count;
//            printf("%d",result);
        }
//        printf("\n");
        count = 0;
    }
    
    printf("%d",result);
    
    return 0;
}