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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}
use std::collections::HashMap;

fn main() {
    input!{
        n: usize,
        v: [i32; n],
    }

    let mut map1 = HashMap::new();
    let mut map2 = HashMap::new();
    // Review some books.
    for i in 0..(n / 2) {
        let num1 = v[i * 2];
        let num2 = v[i * 2 + 1] ;

        // 뮤터블을 만들면 , 다른 참조(임뮤터블 포함해서)를 만들수없음
        if map1.contains_key(&num1) {
            // 인덱스 참조서 임뮤터블 참조를 만든다.
            let t = map1[&num1];
            // 여기서 map1.insert를 통해ㅓ map1를 바꿈. 뮤터블 참조가 내부에서 생김(함수 정의 확인해보면 암.) 그래서 다른 참조를 만들지 못함.
            map1.insert(num1, t + 1);
        } else {
            map1.insert(num1, 1);
        }
        if map2.contains_key(&num2) {
            let t = map2[&num2];
            map2.insert(num2, t + 1);
        } else {
            map2.insert(num2, 1);
        }
    }

    let mut max1 = 0;
    let mut maxk1 = 0;
    let mut max2 = 0;
    let mut maxk2 = 0;
    // Iterate over everything.
    for (k, v) in &map1 {
        if(*v > max1){
            max1 = *v;
            maxk1 = *k;
        }
    }
    for (k, v) in &map2 {
        if(*v > max2){
            max2 = *v;
            maxk2 = *k;
        }
    }
    use std::iter::FromIterator;

    let mut v1 = Vec::from_iter(map1);
    v1.sort_by(|&(_, a), &(_, b)| a.cmp(&b));
    let mut v2 = Vec::from_iter(map2);
    v2.sort_by(|&(_, a), &(_, b)| a.cmp(&b));

    let mut ans = 0;

    if(maxk2 == maxk1){
        let (_, m1) = 
            if v1.len() < 2 {
                (0, 0)
            } else{
                v1[1]
            };

        let (_, m2) = 
            if v2.len() < 2 {
                (0, 0)
            } else{
                v2[1]
            };

        ans = std::cmp::max(max1 + m2, max2 + m1);
    } else{
        ans = max1 + max2
    }


    println!("{}", n - ans);
}