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;
class Program
{
	static void Main(string[] args)
	{
		// 入力フェーズ
				int N = int.Parse(Console.ReadLine());
        string[] input = Console.ReadLine().Split(' ');

				// 偶数番目の数列と奇数番目の数列のそれぞれの数字の登場回数を数える
				Dictionary<int, int> even = new Dictionary<int, int>();
				Dictionary<int, int> odd = new Dictionary<int, int>();
				for(int i=0; i<N; i++)
				{
					int v = int.Parse(input[i]);
					if(i%2 == 0)
					{
						if(even.ContainsKey(v))
						{
							even[v] = even[v] + 1;
						}
						else
						{
							even.Add(v, 1);
						}
					}
					else
					{
						if(odd.ContainsKey(v))
						{
							odd[v] = odd[v] + 1;
						}
						else
						{
							odd.Add(v, 1);
						}
					}
				}

				int cntMax = 0, vMax = 0;
				// evenの最大回数出てくる数字とその回数を調べる
				foreach(var e in even)
				{
					if(e.Value > cntMax)
					{
						cntMax = e.Value;
						vMax = e.Key;
					}
				}

				// 2回目に大きな回数出てくる数字とその回数を調べる
				int cntMax2nd = 0, vMax2nd = 0;
				foreach(var e in even)
				{
					if(e.Key != vMax && e.Value > cntMax2nd)
					{
						cntMax2nd = e.Value;
						vMax2nd = e.Key;
					}
				}

				// oddについて同様に調べる
				int o_cntMax = 0, o_vMax = 0;
				// evenの最大回数出てくる数字とその回数を調べる
				foreach(var e in odd)
				{
					if(e.Value > o_cntMax)
					{
						o_cntMax = e.Value;
						o_vMax = e.Key;
					}
				}

				// 2回目に大きな回数出てくる数字とその回数を調べる
				int o_cntMax2nd = 0, o_vMax2nd = 0;
				foreach(var e in odd)
				{
					if(e.Key != o_vMax && e.Value > o_cntMax2nd)
					{
						o_cntMax2nd = e.Value;
						o_vMax2nd = e.Key;
					}
				}
				int ret = 0;
				// 1番目に大きい同士が違う数字ならOK
				if(vMax != o_vMax)
				{
						ret = N - cntMax - o_cntMax;
				}
				else
				{
					// 1番目と2番目 or 2番目と1番目
					ret = N - Math.Max(cntMax + o_cntMax2nd, cntMax2nd + o_cntMax);
				}
				Console.WriteLine(ret);

    }

		class point
		{
			public int x{set; get;}
			public int y{set; get;}
			public int h{set; get;}
		}
}