Problem:
Score : 300 points

Problem Statement

A sequence a_1,a_2,... ,a_n is said to be /\/\/\/ when the following conditions are satisfied:

For each i = 1,2,..., n-2, a_i = a_{i+2}.

Exactly two different numbers appear in the sequence.

You are given a sequence v_1,v_2,...,v_n whose length is even.
We would like to make this sequence /\/\/\/ by replacing some of its elements.
Find the minimum number of elements that needs to be replaced.

Constraints

2 \leq n \leq 10^5

n is even.

1 \leq v_i \leq 10^5

v_i is an integer.

Input

Input is given from Standard Input in the following format:

n
v_1 v_2 ... v_n

Output

Print the minimum number of elements that needs to be replaced.

Sample Input 1

4
3 1 3 2

Sample Output 1

1

The sequence 3,1,3,2 is not /\/\/\/, but we can make it /\/\/\/ by replacing one of its elements: for example, replace the fourth element to make it 3,1,3,1.

Sample Input 2

6
105 119 105 119 105 119

Sample Output 2

0

The sequence 105,119,105,119,105,119 is /\/\/\/.

Sample Input 3

4
1 1 1 1

Sample Output 3

2

The elements of the sequence 1,1,1,1 are all the same, so it is not /\/\/\/.

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

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::collections::HashMap;

fn main() {
    let n: usize = read();
    let mut even: HashMap<usize, usize> = HashMap::new();
    let mut odd: HashMap<usize, usize> = HashMap::new();
    let mut e1: (usize, usize) = (0, 0);
    let mut e2: (usize, usize) = (0, 0);
    let mut o1: (usize, usize) = (0, 0);
    let mut o2: (usize, usize) = (0, 0);

    for i in 0..n {
        let a: usize = read();
        if i % 2 == 0 {
            let current = even.entry(a).or_insert(0);
            *current += 1;
        } else {
            let current = odd.entry(a).or_insert(0);
            *current += 1;
        }
    }

    for (k, v) in even {
        if v > e1.1 {
            e1 = (k, v);
        } else {
            if v > e2.1 {
                e2 = (k, v);
            }
        }
    }
    for (k, v) in odd {
        if v > o1.1 {
            o1 = (k, v);
        } else {
            if v > o2.1 {
                o2 = (k, v);
            }
        }
    }

    if e1.0 == o1.0 {
        let a = e2.1 + o1.1;
        let b = e1.1 + o2.1;
        if a > b {
            println!("{}", n - a);
        } else {
            println!("{}", n - b);
        }
    } else {
        println!("{}", n - e1.1 - o1.1)
    }
}