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: Accepted)

object Main extends App {
  import scala.io.StdIn._
  val n = readLine.trim.toInt
  val num = readLine.trim.split(' ').map(_.toInt)
  val oddCount = Array.tabulate(100001){_ ⇒ 0}
  val evenCount = Array.tabulate(100001){_ ⇒ 0}
  for (i ← num.indices){
    if (i % 2 == 0){
      evenCount(num(i)) += 1
    }else {
      oddCount(num(i)) += 1
    }
  }
  val oddSorted = oddCount.zipWithIndex.sortBy{case (v, _) ⇒ v}.reverse
  val evenSorted = evenCount.zipWithIndex.sortBy{case (v, _) ⇒ v}.reverse
  println(
    if (oddSorted.head._2 == evenSorted.head._2) {
      if (oddSorted(0)._1 + evenSorted(1)._1 > oddSorted(1)._1 + evenSorted(0)._1) n - (oddSorted(0)._1 + evenSorted(1)._1) else n - (oddSorted(1)._1 + evenSorted(0)._1)
    }else n - (oddSorted.head._1 + evenSorted.head._1)
  )
}