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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main {
  def main(args: Array[String]): Unit = {
    solve
  }

  // 処理の方法はわかるけれど、全然整理できたコードにならない。。
  def solve(): Unit = {
    val sc = new java.util.Scanner(System.in)
    val n = sc.nextInt
    val v = new Array[Int](n)
    for (i <- 0 until n) {
      v(i) = sc.nextInt
    }

    val mp = scala.collection.mutable.Map[Int, Int]()
    for (i <- 0 until n by 2) {
      val num = v(i)
      if (mp.contains(num)) mp(num) += 1
      else mp(num) = 1
    }
    val l1 = mp.toList.sortBy(_._2).reverse

    val mp2 = scala.collection.mutable.Map[Int, Int]()
    for (i <- 1 until n by 2) {
      val num = v(i)
      if (mp2.contains(num)) mp2(num) += 1
      else mp2(num) = 1
    }
    val l2 = mp2.toList.sortBy(_._2).reverse

    if (l1(0)._1 != l2(0)._1) {
      println(n - (l1(0)._2 + l2(0)._2))

    } else if (l1.size == 1 || l2.size == 1) {
      if(l1.size == 1) {
        println(n - l2(0)._2)
      }else{
        println(n - l1(0)._2)
      }
    } else {
      if ((l1(0)._2 - l1(1)._2) >= (l2(0)._2 - l2(1)._2)) {
        // l2の最頻値を残す
        println(n - (l2(0)._2 + l1(1)._2) + l1(0)._2)
      } else {
        // l1の最頻値を残す
        println(n - (l1(0)._2 + l2(1)._2) + l2(0)._2)
      }

    }
  }

}