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)

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;
#ifndef FAST_IO
 #define IOS ios_base::sync_with_stdio(false);cin.tie(nullptr);
#endif

#define FILE_IO freopen("input.txt","r",stdin)
#define int long long
#define f first
#define s second
#define endl '\n'
#define all(c) c.begin(),c.end()
#define debug(x) cerr << #x << " = " << x << endl;

const int MOD=1e9+7;
const int INF=1<<29;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T>void print(T t){for(auto &it:t)cout<<it<<" ";}
inline int powerexp(int a,int b,int m=MOD){int res=1;while(b){if(b&1)res=res*a%m;a=a*a%m,b>>=1;}return res%m;}
inline int modinv(int a){return powerexp(a,MOD-2,MOD);}
inline int lcm(int a,int b){int g=__gcd(a,b);return a/g*b;}

//-------------------< start here >-------------------------//
int calc(string s){
   int n=(int)s.length();
   int ans=-INF;
   for(int i=1;i<n;i++){
	    int cnt=0;
	    for(char c='a';c<='z';c++){
		    bool left=false,right=false;
		    for(int j=0;j<i;j++){if(s[j]==c)left=true;}
		    for(int j=i;j<n;j++){if(s[j]==c)right=true;}
		    if(left && right){
			   cnt++;	
			}	
		 } 
	 ans=max(ans,cnt);
	  }
	return ans;
}
void solve(){
      string s;
      int n;
      cin>>n;
      cin>>s;
      int ret=calc(s);
      cout<<ret<<endl;
}
//-----------------------------
//comment from here for TOPCODER 
int32_t main(){
   IOS;
   solve();
   #ifdef LOCAL_DEFINE
   cerr<<"time elapsed "<<(double)clock()/CLOCKS_PER_SEC<<endl;
   #endif
}