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 = {
    val sc = new java.util.Scanner(System.in)

    // 偶数番目, 奇数番目の数が, それぞれ全て同じであればいい
    // かつ, どちらも異なる数

    val N = sc.nextInt()
    val Vm = 100000
    val OddN, EvenN = new Array[Int](Vm+1)

    (0 until N) foreach { i =>
      val v = sc.nextInt()
      if (i % 2 == 0) {
        EvenN(v) += 1
      } else  {
        OddN(v) += 1
      }
    }

    // Odd(i) // i番目に出現回数が大きい数
    // OddN(Odd(i)) // i番目に出現回数の大きい数の出現回数

    def solve(): Int = {
      val Odd = (0 to Vm).sortBy(i => OddN(i))(Ordering.Int.reverse)
      val Even = (0 to Vm).sortBy(i => EvenN(i))(Ordering.Int.reverse)

      var i = 0
      var j = 0
      while (i < Vm && j < Vm) {
        val oN = OddN(Odd(i))
        val eN = EvenN(Even(j))

        if (oN == 0)
          return N - eN
        else if (eN == 0)
          return N - oN

        if (Odd(i) != Even(j)) {
          return N - oN - eN
        }

        if (Odd(i+1) > Even(i+1))
          i += 1
        else
          j += 1
      }
      N
    }
    println(solve())
  }

}