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> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return pc(48);
  if (x<0) x=-x,pc('-');
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=1e5;
typedef int fuko[yuzu|10];
char c[yuzu|10];
int n=read(),ans;
int main(){
scanf("%s",c+1);
for (int i=1;i<=n;++i){
  int cnt1[999]={0},cnt2[999]={0},tmp=0;
  for (int j=1;j<=i;++j) cnt1[c[j]]++;
  for (int j=i+1;j<=n;++j) if (cnt1[c[j]]&&!cnt2[c[j]]++) ++tmp;
  ans=max(ans,tmp);  
  }write(ans);
}