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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AABC
{
    class Program
    {
        static void Main(string[] args)
        {
            string len = Console.ReadLine();
            string[] nary = Console.ReadLine().Split(' ');

            int ilen = int.Parse(len);
            int[] gBest = new int[100000];
            int[] kBest = new int[100000];

            // (1)偶数奇数要素の一番多い数をカウント
            for(int i = 0; i < ilen; i++)
            {
                // 偶数
                if(i%2 == 0)
                {
                    gBest[i / 2] = int.Parse(nary[i]);
                }
                // 奇数
                else if(i%2 == 1)
                {
                    kBest[(i-1) / 2] = int.Parse(nary[i]);
                }
            }

            int sht = 0;
            int shc = -1;
            for (int p = 0; p < (ilen / 2); p++)
            {
                sht = gBest[p];
                int count = 0;
                for(int q = 0; q < (ilen / 2); q++)
                {
                    if(sht == gBest[q])
                    {
                        shc++;
                    }
                }
                if(count>shc)
                {
                    shc = count;
                }
            }

            int sht2 = 0;
            int shc2 = -1;
            for (int x = 0; x < (ilen / 2); x++)
            {
                sht2 = kBest[x];
                int count = 0;
                for (int y = 0; y < (ilen / 2); y++)
                {
                    if (sht == gBest[y])
                    {
                        shc2++;
                    }
                }
                if (count > shc2)
                {
                    shc2 = count;
                }
            }

            // (2)g,kの一番多い数件数結果から
            int gch = (ilen / 2) - shc;
            int kch = (ilen / 2) - shc2;

            int r = gch + kch;

            int ongcout = 0;
            for (int a = 0; a < ilen; a++)
            {
                if (int.Parse(nary[a]) == int.Parse(nary[0]))
                {
                    ongcout++;
                }
            }
            if(ongcout == ilen)
            {
                r = ilen / 2;
            }

            Console.WriteLine(r);
        }
    }
}