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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <deque>
#include <queue>
using namespace std;
#define rep(i, a, b) for(int i = a; i < (b); ++i)
typedef long long ll;
typedef pair<int,int> P;
int fund(string s, string t){
    int ans = 0;
    for (char c = 'a';c<='z';c++){
        bool isfandA = false;
        bool isfandB = false;
        rep(j, 0, s.length()){
            if (c == s[j]){
                isfandA = true;
            }
        }
        rep(j, 0, s.length()){
            if (c == t[j]){
                isfandB = true;
            }
        }
        if(isfandA && isfandB) ans++;
    }
    return ans;
}
int main() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    int ans = 0;
    
    for(int i=0;i<=n;i++){
        int tmp = fund(s.substr(0,i), s.substr(i));
        ans = max(ans, tmp);
    }
    
    cout << ans << endl;
    return 0;
}