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.Threading.Tasks;

namespace ARC103C
{
    class Program
    {
        struct Count
        {
            public int Number;
            public int Counts;
        }
        static void Main(string[] args)
        {
            var n = int.Parse(Console.ReadLine());
            int[] vn = Console.ReadLine().Split().Select(int.Parse).ToArray();

            //aの最頻値，bの最頻値を求める
            Count[] a = new Count[100001];
            Count[] b = new Count[100001];
            for(int i = 0; i < 100000; i++)
            {
                a[i].Number = i;
                b[i].Number = i;
            }
            for(int i = 0; i < n; i++)
            {
                if (i % 2 == 0) a[vn[i]].Counts++;
                else b[vn[i]].Counts++;
            }

            //max(aの最頻値+bの最頻値,aの最頻値＋bの第二最頻値，bの最頻値+aの第二最頻値)採用
            a = a.OrderByDescending(x => x.Counts).ToArray();
            b = b.OrderByDescending(x => x.Counts).ToArray();
            int amax = a[0].Counts;
            int amax2 = a[1].Counts;
            int bmax = b[0].Counts;
            int bmax2 = b[1].Counts;

            int minus = amax + bmax;
            if (a[0].Number == b[0].Number) minus = Math.Max(amax + bmax2, amax2 + bmax);

            int ans = n - minus;
            Console.WriteLine(ans);
            Console.ReadKey();
        }
    }
}