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.Linq;
using System.Collections.Generic;

namespace Algorithm
{
    class Program
    {
        static void Main(string[] args)
        {
            var n = int.Parse(Console.ReadLine());
            var array = Console.ReadLine().Split().Select(z => int.Parse(z)).ToArray();

            var oddDict = new Dictionary<int, int>();
            var eveDict = new Dictionary<int, int>();

            for (var i = 0; i < n; i++)
            {
                if (i % 2 == 0)
                    oddDict[array[i]] = 0;
                else
                    eveDict[array[i]] = 0;
            }

            for (var i = 0; i < n; i++)
            {
                if (i % 2 == 0)
                    oddDict[array[i]]++;
                else
                    eveDict[array[i]]++;
            }

            var oddMax = oddDict.First(x => x.Value == oddDict.Max(m => m.Value)).Key;
            var eveMax = eveDict.First(x => x.Value == eveDict.Max(m => m.Value)).Key;

            // oddとeveが同じ場合は特殊処理
            if (oddMax == eveMax)
            {
                Console.WriteLine(n / 2);
                return;
            }

            // oddMaxとeveMaxではない数値を数える
            var total = 0;

            for (var i = 0; i < n; i++)
            {
                if (i % 2 == 0 && array[i] != oddMax)
                    total++;

                if (i % 2 == 1 && array[i] != eveMax)
                    total++;
            }

            Console.WriteLine(total);
        }
    }
}