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.Collections.Generic;
using System.Linq;

public static class Program
{
    public static void Main(string[] args)
    {
        var n = int.Parse(Console.ReadLine());
        var vals = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToList();
        var tmp = vals.Distinct().ToList();
        if (tmp.Count == 1)
            Console.WriteLine(Math.Floor((double)n / 2));
        else
        {
            var evenDict = new Dictionary<int, int>();
            var oddDict = new Dictionary<int, int>();
            for (int i = 0; i < n; i++)
            {
                if (i % 2 == 0)
                {
                    if (evenDict.ContainsKey(vals[i]))
                        evenDict[vals[i]] += 1;
                    else
                        evenDict[vals[i]] = 1;
                }
                else
                {
                    if (oddDict.ContainsKey(vals[i]))
                        oddDict[vals[i]] += 1;
                    else
                        oddDict[vals[i]] = 1;
                }
            }
            var evenMaxKey = evenDict.OrderByDescending(val => val.Value).First().Key;
            var oddMaxKey = oddDict.OrderByDescending(val => val.Value).First().Key;
            var evenCnt = evenDict.Select(x => x.Value).Sum() - evenDict[evenMaxKey];
            var oddCnt = oddDict.Select(x => x.Value).Sum() - oddDict[oddMaxKey];
            Console.WriteLine(evenCnt + oddCnt);
        }
    }
}