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::stdin;
use std::time::SystemTime;
use std::collections::HashMap;

#[allow(dead_code)]
fn read_line() -> String {
    let mut s = String::new();
    stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

#[allow(dead_code)]
fn read<T>() -> T
where T: std::str::FromStr
{
    read_line().parse::<T>().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T>() -> Vec<T>
where T: std::str::FromStr
{
    read_line().split(" ").map(|n| n.parse::<T>().ok().unwrap()).collect()
}

#[allow(dead_code)]
fn read_chars() -> Vec<char>
{
    read_line().chars().collect()
}

#[allow(dead_code)]
fn print_time() -> () {
    println!("{:?}", SystemTime::now());
}

#[allow(unused_variables)]
fn main() {
    let n = read::<u64>();
    let vec: Vec<u64> = read_vec();
    let mut first_map = HashMap::new();
    let mut second_map = HashMap::new();
    
    let mut f = 0;
    let mut s = 0;
    let mut l = true;

    for (i, v) in vec.into_iter().enumerate() {
        if i % 2 == 0 {
            let c = first_map.get(&v).cloned().unwrap_or(0);
            first_map.insert(v, c + 1);
            if s == 0 {
                s = v;
            } else if s != v {
                l = false;
            }
        } else {
            let c = second_map.get(&v).cloned().unwrap_or(0);
            second_map.insert(v, c + 1);
            if f == 0 {
                f = v;
            } else if f != v {
                l = false;
            }
        }
    }
    if l {
        l = f != s;
    }
    
    if l {
        println!("0");
    } else {
        match (first_map.len(), second_map.len()) {
            (1, 1) => println!("{}", n / 2),
            _ => {
                let mut first_num = 0;
                let mut first_num2 = 0;
                let mut first_max = 0;
                let mut first_max2 = 0;
                for (k, v) in first_map {
                    if first_max < v {
                        first_num = k;
                        first_max = v;
                    } else if first_num2 < v {
                        first_num2 = k;
                        first_max2 = v;
                    }
                }
            
                let mut second_num = 0;
                let mut second_num2 = 0;
                let mut second_max = 0;
                let mut second_max2 = 0;
                for (k, v) in second_map {
                    if second_max < v {
                        second_num = k;
                        second_max = v;
                    } else if second_num2 < v {
                        second_num2 = k;
                        second_max2 = v;
                    }
                }
                
                if first_num == second_num {
                    if (first_max + second_max2) < (first_max2 + second_max) {
                        println!("{}", n - first_max2 - second_max);
                    } else {
                        println!("{}", n - first_max - second_max2);
                    }
                } else {
                    println!("{}", n - first_max - second_max);
                }
            }
        }
    }
}