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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using E = System.Linq.Enumerable;
 
class Program
 {
        static void Main(string[] args)
        {
          //入力を受け取る
          var n = int.Parse(Console.ReadLine());
          var v = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
          var Odd  = new Dictionary<int, int>();
          var Even = new Dictionary<int, int>();
          
          for(int i = 0; i < n ; i++){
            if(i%2 == 0){
              if(Even.ContainsKey(v[i])){
                Even[v[i]] += 1;
              }else{
                Even.Add(v[i],1);
              }
            }else{
              if(Odd.ContainsKey(v[i])){
                Odd[v[i]] += 1;
              }else{
                Odd.Add(v[i],1);
              }
            }
          }
          
          int maxOddValue=0;
          int maxOddKey=0;
          int SecondOddValue=0;
          int SecondOddKey=0;
          
          foreach(KeyValuePair<int, int> item in Odd) {
            if(maxOddValue < item.Value){
              SecondOddValue = maxOddValue;
              SecondOddKey   = maxOddKey;
              maxOddValue = item.Value;
              maxOddKey   = item.Key;
            }else if(SecondOddValue < item.Value){
              SecondOddValue = item.Value;
              SecondOddKey   = item.Key;
            }
          }
          
          int maxEvenValue=0;
          int maxEvenKey=0;
          int SecondEvenValue=0;
          int SecondEvenKey=0;
          
          foreach(KeyValuePair<int, int> item in Even) {
            if(maxEvenValue < item.Value){
              SecondEvenValue = maxEvenValue;
              SecondEvenKey   = maxEvenKey;
              maxEvenValue = item.Value;
              maxEvenKey   = item.Key;
            }else if(SecondEvenValue < item.Value){
              SecondEvenValue = item.Value;
              SecondEvenKey   = item.Key;
            }
          }
          
          //それぞれの変更に必要な回数を求める
          int changeOdd = (n/2)-maxOddValue;
          int changeEven = (n/2)-maxEvenValue;
          
          if(maxOddKey == maxEvenKey){
              changeOdd = (n/2)-SecondOddValue;
              var a = changeOdd+changeEven;
              changeOdd = (n/2)-maxOddValue;
              changeEven = (n/2)-SecondEvenValue;
              var b = changeOdd+changeEven;
              
            if(a < b){
              Console.WriteLine(a);
            }else{
              Console.WriteLine(b);
            }
          }else{
            Console.WriteLine(changeOdd+changeEven);
          }
        }
}