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;

class Program {
    static void Main() {
        Scanner cin = new Scanner();
        int N = cin.Int();
        var v = cin.ArrayInt(N);

        var odd = new List<int>();
        var even = new List<int>();
        for (int i = 0; i <= N - 2; i += 2) {
            odd.Add(v[i]);
        }
        for(int i = 1; i <= N - 1; i += 2) {
            even.Add(v[i]);
        }

        var odd_dic = new SortedDictionary<int, int>();
        odd_dic.Add(int.MaxValue, 0);
        var even_dic = new SortedDictionary<int, int>();
        even_dic.Add(int.MaxValue, 0);
        foreach (var x in odd) {
            if (odd_dic.ContainsKey(x)) odd_dic[x]++;
            else odd_dic.Add(x, 1);
        }
        foreach (var x in even) {
            if (even_dic.ContainsKey(x)) even_dic[x]++;
            else even_dic.Add(x, 1);
        }

        if(odd_dic.Keys.ElementAt(0) != even_dic.Keys.ElementAt(0)) {
            Console.WriteLine(N - odd_dic.Values.ElementAt(0) - even_dic.Values.ElementAt(0));
        }
        else {
            Console.WriteLine(Math.Min(N - odd_dic.Values.ElementAt(0) - even_dic.Values.ElementAt(1), N - odd_dic.Values.ElementAt(1) - even_dic.Values.ElementAt(0)));
        }
        Console.ReadLine();
    }
}

class Scanner //標準入力
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int Int()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = Int() + add;
        }
        return Array;
    }

    public long Long()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = Long() + add;
        }
        return Array;
    }

    public double Double()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = Double() + add;
        }
        return Array;
    }
}