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

using System.Collections;

class Program
{
    static void Main()
    {
        int n = int.Parse(Console.ReadLine());

        var v = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();

        int[] v1 = new int[n / 2];
        int[] v2 = new int[n / 2];

        for (int i = 0; i < n; i++)
        {
            var val = v[i];
            
            if(i % 2 == 0)
            {
                v1[i / 2] = val;
            }
            else
            {
                v2[i / 2] = val;
            }
        }

        //v1の中で最も多い数値と2番目に多い数値
        var xx1 = v1.GroupBy(val => val).OrderByDescending(grp => grp.Count()).Take(2);
        var xx2 = v2.GroupBy(val => val).OrderByDescending(grp => grp.Count()).Take(2);

        if(xx1.First().Key != xx2.First().Key)
        {
            //それぞれ最も多い数で統一させれば良い
            Console.WriteLine(n - xx2.First().Count() - xx1.First().Count());
        }
        else
        {
            var val1 = n - xx1.First().Count() - (xx2.Count() == 2 ? xx2.Skip(1).First().Count() : 0);
            var val2 = n - xx2.First().Count() - (xx1.Count() == 2 ? xx1.Skip(1).First().Count() : 0);

            Console.WriteLine(Math.Min(val1, val2));
        }

    }
    
    
}