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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <iomanip>
#include <math.h>
//#include <unistd.h>
//#include <stdio.h>
#define rep(i,x) for(ll i = 0;i<x;i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
using dou = double;
const ll INF = 1LL << 60;
const ll mod = 1000000007;
typedef pair<ll,ll> P;
using graph = vector<vector<ll>>;
template<class T> inline bool chmax(T& a, T b){if(a < b){a = b; return true;}return false;}
template<class T> inline bool chmin(T& a, T b){if(a > b){a = b; return true;}return false;}
ll gcd(ll a,ll b){
    if(a < b)swap(a,b);
    if(a % b == 0)return b;
    else return gcd(b,a%b);
}
//素数判定O(sqrt(N))
bool isp(ll n){
    bool res = true;
    if(n == 1)return false;
    else{
        for(ll i = 2;i * i <= n;i++){
            if(n % i == 0){
                res = false;
                break;
            }
        }
        return res;
    }
}

ll fac[510000], finv[510000], inv[510000];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < 510000; i++){
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = mod - inv[mod%i] * (mod / i) % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}

// 二項係数計算
ll COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}
//nのm乗をMODで割ったあまりO(logN)
ll modpow(ll n,ll m,ll MOD){
    ll res = 1;
    while(m){
        if(m & 1)res = (res * n) % MOD;
        m >>= 1;
        n *= n;
        n %= MOD;
    }
    return res;
}

//printf("%.10f\n", N);
/*  vector<ll> vec = {1,2,5,6,8,9,10};
    cout << binary_search(all(vec),5) << endl; -> true*/
/*
    vector<ll> vec = {1,2,5,6,8,9,10};
    auto n = upper_bound(all(vec),4);
    cout << *n << endl; -> 5*/
/*
    cout << fixed << setprecision(15);
*/

int main(){
    ll N;
    string S;
    cin >> N >> S;
    ll ans = 0;
    for(ll i = 1;i <= N - 1;i++){
        string l = S.substr(0,i),r = S.substr(i,N - i);
        map<char,ll> lem;
        map<char,ll> rim;
        rep(j,i){
            lem[l[j]]++;
        }
        rep(j,N - i){
            rim[r[j]]++;
        }
        ll cnt = 0;
        for(auto x : lem){
            if(rim[x.first] != 0){
                cnt++;
            }
        }
        ans = max(ans,cnt);
    }
    cout << ans << endl;
}