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.IO;
using System.Text;
using static System.Console;
using static System.Math;

namespace AtCodeeeer
{
    class AtCOOOOOOOOOOOder
    {
        static void Main(string[] args)
        {
            int n = int.Parse(ReadLine());
            int[] s = ReadLine().Split(' ').Select(int.Parse).ToArray();
            var leftDic = new Dictionary<int, int>();
            var rightDic = new Dictionary<int, int>();
            var All = new Dictionary<int, int>();

            for (int i = 0; i < n; i++)
            {
                //left
                if(i%2 == 0)
                {
                    if (!leftDic.ContainsKey(s[i])) leftDic[s[i]] = 1;
                    else{
                        leftDic[s[i]]++;
                    }
                }
                else
                {
                    if (!rightDic.ContainsKey(s[i])) rightDic[s[i]] = 1;
                    else
                    {
                       rightDic[s[i]]++;
                    }
                }

                if (!All.ContainsKey(s[i])) All[s[i]] = 1;
                else
                {
                    All[s[i]]++;
                }
            }
            var sortedLeftDic = leftDic.OrderByDescending(x => x.Value).ToArray();
            var sortedRightDic = rightDic.OrderByDescending(x => x.Value).ToArray();



            //判定
            if(All.Count == 1) { WriteLine(n / 2);return;}

            if (sortedLeftDic[0].Key == sortedRightDic[0].Key)
            {
                WriteLine(
                    n - (Max(sortedLeftDic[0].Value, sortedRightDic[0].Value) + Max(sortedLeftDic[1].Value, sortedRightDic[1].Value))
                    );
            }else
                WriteLine(
                    n - (sortedLeftDic[0].Value + sortedRightDic[0].Value)
                    );

        }
    }
}