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>
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define ll long long
using namespace std;

const ll P = 1000000007;
int gcd(int a, int b) { return b != 0 ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
vector<char> Alphabet = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
vector<char> alphabet = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};

int main()
{
  cout << fixed << setprecision(10);
  int N;
  cin >> N;
  string S;
  cin >> S;
  int ans = 0;
  rep(i,N)
  {
    int cnt = 0;
    set<char> se;
    set<char> se2;
    for(int j=0;j<N;j++)
    {
      if(i<=j)
      {
        se.insert(S[j]);
      }
      else
      {
        se2.insert(S[j]);
      }
    }
    rep(j,26)
    {
      if(se.count(alphabet[j]))
      {
        if(se2.count(alphabet[j])) 
        {
          cnt ++;
        }
      }
    }
    ans = max(ans,cnt);
  }
  cout << ans << endl;
  return 0;
}