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 PHP (5.6.30) (judge verdict: Wrong Answer)

<?php
$n = trim(fgets(STDIN));
$v = explode(" ", trim(fgets(STDIN)));
$times = 0;

for($i = 0; $i < count($v); $i+=2){
	$odd[] = $v[$i];
}
for($i = 1; $i < count($v); $i+=2){
	$even[] = $v[$i];
}
//$count = array_count_values($v);
$odd_count = array_count_values($odd);
$even_count = array_count_values($even);
arsort($odd_count);
arsort($even_count);
$odd_keys = array_keys($odd_count);
$even_keys = array_keys($even_count);
$odd_values = array_values($odd_count);
$even_values = array_values($even_count);

/*var_dump($odd_count);
var_dump($even_count);
var_dump($odd_keys);
var_dump($even_keys);
var_dump($odd_values);
var_dump($even_values);
var_dump($odd_keys[0]);
var_dump($even_keys[0]);*/

//一番出現数の多い数値が同じか確認
//同じじゃないなら2種類が担保されているから普通にいくつ入れ替えるか確認する
if($odd_keys[0] != $even_keys[0]){
	//奇数番目の数の種類が1よりも多いなら入れ替え処理を行う
	if(count($odd_values) > 1){
		$times+=(count($odd) - $odd_values[0]);
	}
	//偶数番目の数の種類が1よりも多いなら入れ替え処理を行う
	if(count($even_values) > 1){
		$times+=(count($even) - $even_values[0]);
	}
}else{
	//一番出現数の多い数値が同じなので奇数番目と偶数番目でどちらが一番出現数の多い数値に合わせて入れ替えるか確認する
	//入れ替え回数の少ない番目は一番出現数の多い数値に合わせて入れ替える
	//奇数番目の方が入れ替え回数が多いので奇数番目は二番目に出現数の多い数値に合わせて入れ替える
	if(count($odd) - $odd_values[0] > count($even) - $even_values[0]){
		if(count($odd_values) > 1){
			$times+=(count($odd) - $odd_values[1]);
		}
		if(count($even_values) > 1){
			$times+=(count($even) - $even_values[0]);
		}
	}else{
		if(count($odd_values) > 1){
			$times+=(count($odd) - $odd_values[0]);
		}
		if(count($even_values) > 1){
			$times+=(count($even) - $even_values[1]);
		}
	}
	if($times == 0){
		$times = 2;
	}
}

printf("%d", $times);
?>