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 (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;
use std::collections::BTreeMap;
use std::cmp::Reverse;

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

    let mut odds = BTreeMap::new();
    let mut evens = BTreeMap::new();

    for (i, v) in vv.into_iter().enumerate() {
        if i % 2 == 1 {
            *odds.entry(Reverse(v)).or_insert(0) += 1;
        } else {
            *evens.entry(Reverse(v)).or_insert(0) += 1;
        }
    }
 
    if let (Some((&Reverse(o_key), &o_val)), Some((&Reverse(e_key), &e_val)))
        = (odds.iter().next(), evens.iter().next()) 
        {
            if o_key != e_key {
                let ans = n - o_val - e_val;
                println!("{}", ans);
            } else if let (Some((&Reverse(_), &o2_val)), Some((&Reverse(_), &e2_val)))
                = (odds.iter().nth(1), evens.iter().nth(1)) 
            {
                let ans = n - std::cmp::max(o_val + e2_val, o2_val + e_val);
                println!("{}", ans);
            } else {
                let ans = n - o_val;
                println!("{}", ans);
            }
    }

}