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)

#define _CRT_SECURE_NO_WARNINGS

/* include ***********************/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

/* define *************************/
//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)
#define FOR(i,s,n) for(int i=(int)(s),i##_len=(int)(n);i<i##_len;i++)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTL(d) printf("%lld\n",(d))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

#define INF 1e9

typedef long long ll;


int diff[4][2] = {
	{ 0, -1 },
	{ -1, 0 },
	{ 1, 0 },
	{ 0, 1 },
};

int a[200000];

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int a, int b) {
	return (a) > (b) ? (a) : (b);
}
void Swap(int *a, int *b) {
	int tmp = *a;
	*a = *b;
	*b = tmp;
}

int main() {
	int n;
	char s[101];
	int select[26] = { 0 };
	int cnt;
	int max = 0;

	scanf("%d", &n);
	scanf("%s", s);

	REPS(i, n) {
		cnt = 0;
		m0(select);
		REP(j, n) {
			if (j < i) {
				if (select[s[j] - 'a'] == 0)select[s[j] - 'a']++;
			}
			else {
				if (select[s[j] - 'a'] == 1)cnt++;
				select[s[j] - 'a']+=2;
			}
		}
		//PRINTD(cnt);
		if (max < cnt)max = cnt;
	}
	PRINTD(max);
}