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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define PI 3.1415926535897932
#define MOD 1000000007
#define INF 1e9

#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define FOR(i, start, end) for(int i = (start); i < (end); i++)
#define ALL(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef vector<bool> vb;


void solve()
{
    int N; cin >> N;
    string S; cin >> S;

    int ans = 0;
    for(int i = 1; i < N-1; i++)
    {
        string X = S.substr(0, i);
        string Y = S.substr(i, N);
        int countX = 0, countY = 0;
        REP(i, X.length()) countX |= 1 << X[i]-'a';
        REP(i, Y.length()) countY |= 1 << Y[i]-'a';
        ans = max(ans, (int)bitset<32>(countX & countY).count());
    }
    cout << ans << endl;
}

int main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    solve();

    return 0;
}