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: Wrong Answer)

#include <stdio.h>

int main ()
{
    int N = 0;
    scanf("%d", &N);

    char S[N];
    scanf("%s", S);

    int ans = 0;
    int counter = 0;
    char X[N], Y[N];

    X[0] = '\0';
    Y[0] = '\0';
    
    for (int i = 0; i < N; i++)
    {
        counter = 0;
        for (int j = 0; j <= i; j++)
        {
            X[j] = S[j];
        }
        for (int k = i; k < N; k++)
        {
            Y[k - i] = S[k + 1];
        }
        
        for (int L = 0; L < N; L++)
        {
            for (int M = L + 1; M < N; M++)
            {
                if (X[L] == X[M])
                {
                    X[M] = '\0';
                }
                if (Y[L] == Y[M])
                {
                    Y[M] = '\0';
                }
            }
        }

        for (int L = 0; L < N; L++)
        {
            for (int M = 0; M < N; M++)
            {
                if (X[L] == '\0' || Y[M] == '\0')
                {
                }
                else if (X[L] == Y[M])
                {
                    counter += 1;
                }
            }
        }

        if (counter > ans)
        {
            ans = counter;
        }
    }
    printf("%d", ans);
    return 0;
}