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)

#![allow(non_snake_case)]

fn read_line() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read_line()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let n: usize = read_line().parse().unwrap();
    let v: Vec<i32> = read_vec();
    let even: Vec<i32> = (0..n/2).map(|i| v[i*2]).collect();
    let odd: Vec<i32> = (0..n/2).map(|i| v[i*2+1]).collect();
    let mut hash = std::collections::HashMap::new();
    for i in even.iter() {
        let count = hash.entry(i).or_insert(0);
        *count += 1;
    }
    let mut even_max = (0, 0);
    let mut even_max2 = (0, 0);
    for (&&k, &v) in hash.iter() {
        if v > even_max.1 {
            even_max2 = even_max;
            even_max = (k, v);
        } else if v > even_max2.1 {
            even_max2 = (k, v);
        }
    }
    hash.clear();
    for i in odd.iter() {
        let count = hash.entry(i).or_insert(0);
        *count += 1;
    }
    let mut odd_max = (0, 0);
    let mut odd_max2 = (0, 0);
    for (&&k, &v) in hash.iter() {
        if v > odd_max.1 {
            odd_max2 = odd_max;
            odd_max = (k, v);
        } else if v > odd_max2.1 {
            odd_max2 = (k, v);
        }
    }
    let ans = if even_max.0 != odd_max.0 {
        n - even_max.1 - odd_max.1
    } else if even_max.1 > odd_max.1 {
        n - even_max.1 - odd_max2.1
    } else if even_max.1 > odd_max.1 {
        n - even_max2.1 - odd_max.1
    } else if even_max2.1 > even_max2.1 {
        n - even_max2.1 - odd_max.1
    } else {
        n - even_max.1 - odd_max2.1
    };
    println!("{}", ans);
}