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::Read;
use std::collections::HashMap;
use std::cmp::min;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();

    let mut counts_b: HashMap<usize, usize> = HashMap::new();
    let mut counts_c: HashMap<usize, usize> = HashMap::new();
    for i in 0..n {
        let ai = iterator.next().unwrap().parse().unwrap();
        if i % 2 == 0 {
            *counts_b.entry(ai).or_insert(0) += 1;
        } else {
            *counts_c.entry(ai).or_insert(0) += 1;
        }
    }

    let mut b = 0;
    let mut max_count_b = 0;
    for (&x, &count) in counts_b.iter() {
        if count > max_count_b {
            b = x;
            max_count_b = count;
        }
    }

    let mut c = 0;
    let mut max_count_c = 0;
    for (&x, &count) in counts_c.iter() {
        if count > max_count_c {
            c = x;
            max_count_c = count;
        }
    }

    if b != c {
        let ans = n - max_count_b - max_count_c;
        return ans.to_string();
    }

    let mut second_max_count_b = 0;
    for (&x, &count) in counts_b.iter() {
        if x == b {
            continue;
        }
        if count > second_max_count_b {
            b = x;
            second_max_count_b = count;
        }
    }

    let mut second_max_count_c = 0;
    for (&x, &count) in counts_c.iter() {
        if x == c {
            continue;
        }
        if count > second_max_count_c {
            c = x;
            second_max_count_c = count;
        }
    }

    let ans = min(
        n - second_max_count_b - max_count_c,
        n - max_count_b - second_max_count_c
    );

    return ans.to_string();
}