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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

pub fn read<t: FromStr>() -> t {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

use std::cmp::max;

fn main() {
    let n: usize = read();
    let s: Vec<char> = read::<String>().chars().collect();
    let mut maxv = 0;

    for i in 1..n - 1 {
        let mut acnt: Vec<i64> = vec![0; 27];
        let mut bcnt: Vec<i64> = vec![0; 27];
        for a in 0..i {
            acnt[(s[a] as u32 - 'a' as u32) as usize] += 1;
        }
        for b in i..n {
            bcnt[(s[b] as u32 - 'a' as u32) as usize] += 1;
        }

        let mut sum = 0;
        for i in 0..27 {
            if bcnt[i] != 0 && acnt[i] != 0 {
                sum += 1;
            }
        }

        maxv = max(maxv, sum);
    }
    println!("{}", maxv);
}