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: Accepted)

use std::cmp::min;
use std::collections::HashMap;
use std::io::prelude::*;

fn main() {
    let stdin = std::io::stdin();
    let mut reader = std::io::BufReader::new(stdin.lock());
    let mut s = String::new();
    reader.read_line(&mut s).unwrap();
    let n: usize = s.trim().parse().unwrap();

    let mut s = String::new();
    reader.read_line(&mut s).unwrap();
    let v: Vec<usize> = s.split_whitespace().map(|x| x.parse().unwrap()).collect();

    let mut odd = HashMap::new();
    let mut even = HashMap::new();

    for i in 0..n {
        if i % 2 == 0 {
            let count = odd.entry(v[i]).or_insert(0);
            *count += 1;
        } else {
            let count = even.entry(v[i]).or_insert(0);
            *count += 1;
        }
    }

    let mut odd: Vec<_> = odd.iter().collect();
    let mut even: Vec<_> = even.iter().collect();
    odd.sort_by_key(|k| k.1);
    even.sort_by_key(|k| k.1);
    odd.reverse();
    even.reverse();

    println!(
        "{}",
        if odd[0].0 != even[0].0 {
            v.len() - odd[0].1 - even[0].1
        } else {
            min(n - odd[0].1 - if even.len() > 1 { *even[1].1 } else { 0 }, n - even[0].1 - if odd.len() > 1 { *odd[1].1 } else { 0 })
        }
    );
}