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;

class Program
{
    static void Main(string[] args)
    {
        var n = int.Parse(Console.ReadLine());
        var v = new int[n];
        var inputs = Console.ReadLine().Split(' ');
        var oddCounts = new Dictionary<int, int>();
        var evenCounts = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
        {
            v[i] = int.Parse(inputs[i]);
            Dictionary<int, int> targetCounts = null;
            if (i % 2 == 0)
            {
                targetCounts = evenCounts;
            }
            else
            {
                targetCounts = oddCounts;
            }

            if (targetCounts.ContainsKey(v[i]))
            {
                targetCounts[v[i]]++;
            }
            else
            {
                targetCounts.Add(v[i], 1);
            }
        }

        var evenTop2 = evenCounts.OrderByDescending((pair) => pair.Value).Take(2).Select(pair => pair.Key).ToList();
        var oddTop2 = oddCounts.OrderByDescending((pair) => pair.Value).Take(2).Select(pair => pair.Key).ToList();
        if (oddTop2.Count == 1)
        {
            oddTop2.Add(0);
        }

        var count = dfs(v, evenTop2, oddTop2);
        Console.WriteLine(count);
    }

    static int dfs(int[] v, List<int> evenTop2, List<int> oddTop2)
    {
        int min = 1000000;
        foreach (var even in evenTop2)
        {
            foreach (var odd in oddTop2)
            {
                if (even == odd)
                {
                    break;
                }

                int count = 0;
                for (int i = 0; i < v.Length; i++)
                {
                    if (i % 2 == 0)
                    {
                        if (v[i] != even)
                        {
                            count++;
                        }
                    }
                    else
                    {
                        if (v[i] != odd)
                        {
                            count++;
                        }
                    }
                }
                min = Math.Min(min, count);
            }
        }
        if (min == 1000000)
        {
            return 2;
        }
        return min;
    }

}