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::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    token.parse().ok().unwrap()
}

fn main() {
    let n = input::<usize>();
    let (v0, v1) = {
        let mut v0 = vec![0; n / 2];
        let mut v1 = vec![0; n / 2];
        for i in 0..n {
            if i % 2 == 0 {
                v0[i / 2] = input::<i64>();
            } else {
                v1[i / 2] = input::<i64>();
            }
        }
        v0.sort();
        v1.sort();
        (v0, v1)
    };

    let mut u0 = vec![];
    let mut u1 = vec![];
    let mut c0 = 1;
    let mut c1 = 1;
    for i in 1..n / 2 {
        if v0[i] == v0[i - 1] {
            c0 += 1;
        } else {
            u0.push((c0, v0[i - 1]));
            c0 = 1;
        }
        if v1[i] == v1[i - 1] {
            c1 += 1;
        } else {
            u1.push((c1, v1[i - 1]));
            c1 = 1;
        }
    }
    if c0 > 0 {
        u0.push((c0, v0[n / 2 - 1]));
    }
    if c1 > 0 {
        u1.push((c1, v1[n / 2 - 1]));
    }
    u0.sort();
    u0.reverse();
    u1.sort();
    u1.reverse();

    let ans = if u0[0].1 == u1[0].1 {
        if u0.len() == 1 && u1.len() == 1 {
            n / 2
        } else if u0.len() == 1 {
            n / 2 - u1[1].0
        } else if u1.len() == 1 {
            n / 2 - u0[1].0
        } else {
            std::cmp::min(n / 2 - u0[1].0, n / 2 - u1[1].0)
        }
    } else {
        n - u0[0].0 - u1[0].0
    };
    println!("{}", ans);
}