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.Text;
using System.Collections.Generic;
using System.ComponentModel;
using System.Linq;
using System.Threading.Tasks;
class Program
{
	static void Main(string[] args)
	{
		int n = int.Parse(Console.ReadLine());
		var input = Console.ReadLine().Split().Select(int.Parse).ToArray();
		List<int> even = new List<int>();
		List<int> odd = new List<int>();
        for(var i = 0; i < n; i++)
		{
			if (i % 2 == 0)
			{
				even.Add(input[i]);
			}
			else
			{
				odd.Add(input[i]);
			}
		}

		even.Sort();
		odd.Sort();
		int[] cnteven = new int[n / 2];
		int[] cntodd = new int[n / 2];
		for(int i = 0; i < n / 2; i++)
		{
			cnteven[i] = 0;
			cntodd[i] = 0;
		}
		int nume = 0;
		int numo = 0;
        for(int i = 0; i < n / 2; i++)
		{
			if (i > 0 && even[i] == even[i - 1])
			{
				cnteven[nume]++;
			}
			else
			{
				if (i > 0) { nume++; }
				cnteven[nume]++;
			}

			if (i > 0 && odd[i] == odd[i - 1])
			{
				cntodd[numo]++;
			}
			else
			{
				if (i > 0) { numo++; }
				cntodd[numo]++;
			}
		}

		Array.Sort(cnteven);
		Array.Sort(cntodd);


		Console.WriteLine(even.Count + odd.Count - cnteven[cnteven.Length - 1] - cntodd[cntodd.Length - 1]);

	}
}