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, unused)]
use std::cmp::*;
use std::collections::*;
use std::f64::consts::*;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input!{
        n: usize,
        mut v: [u64; n],
    }
    let mut even = BTreeMap::new();
    let mut odd = BTreeMap::new();

    for (i, v_) in v.iter().enumerate() {
        if i % 2 == 0 {
            *even.entry(v_).or_insert(0) += 1;
        } else if i % 2 != 0 {
            *odd.entry(v_).or_insert(0) += 1;
        }
    }

    let mut ans = v.len();
    let mut even_ans = v.len();
    let mut odd_ans = v.len();
    let mut flag = 0;
    for ((k_even, val_even), (k_odd, val_odd)) in even.iter().rev().zip(odd.iter().rev()) {
        if k_even != k_odd {
            flag = 1;
            break;
        } else {
            break;
        }
    }

    if flag == 1 {
        for (i, (_, val)) in even.iter().rev().enumerate() {
            if i == 0 {
                ans -= *val;
                break;
            }
        }
        for (i, (_, val)) in odd.iter().rev().enumerate() {
            if i == 0 {
                ans -= *val;
                break;
            }
        }
    } else {
        for (i, (_, val)) in even.iter().rev().enumerate() {
            if i == 0 {
                even_ans -= *val;
                break;
            }
        }
        for (i, (_, val)) in odd.iter().rev().enumerate() {
            if i == 1 {
                even_ans -= *val;
                break;
            }
        }
        for (i, (_, val)) in even.iter().rev().enumerate() {
            if i == 1 {
                odd_ans -= *val;
                break;
            }
        }
        for (i, (_, val)) in odd.iter().rev().enumerate() {
            if i == 0 {
                odd_ans -= *val;
                break;
            }
        }
        ans = min(even_ans, odd_ans);
    }

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