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(void){
    char string[101];
    int num_string;
    int i,j,k=0;
    char X[100],Y[100];
    int kinds=0;
    int count=0,max=0;
    //printf("Input the length of string:");
    scanf("%d",&num_string);
    //printf("Input string:");
    scanf("%s",string);
 
    for(i=0;i<num_string-1;i++){
        //printf("i:%d ",i);
        for(j=0;j<=i;j++){
            X[j]=string[j];
        }
        //printf("X:%s\n",X);
        k=0;
        for(j=i+1;j<num_string;j++){
            Y[k]=string[j];
            //printf("Y[%d]:%c\n",k,Y[k]);
            for(int m=0;m<=i;m++){
                if(Y[k]==X[m]){
                    count++;
                    //printf("count:%d\n",count);
                    for(int n=m;n<=i;n++){
                        //printf("n:%d\n",n);
                        if(Y[k]==X[n]){
                            X[n]=' ';
                        }
                    }
                }
            }
            k++;
            if(kinds<count){
            kinds=count;
            }
        }
        count=0;
 
    }
    printf("%d",kinds);
    return 0;
}