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;

public class Hello
{
    public static void Main()
    {
        // Your code here!
        int n = int.Parse(Console.ReadLine());
        int[] a = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
        
        Dictionary<int, int> odd = new Dictionary<int, int>();
        Dictionary<int, int> even = new Dictionary<int, int>();
        
        for(int i = 0; i < n ; i++)
        {
            if(i % 2 == 0)
            {
                even[a[i]] = 0;
            }
            else
            {
                odd[a[i]] = 0;
            }
            
        }
        
        for(int i = 0; i < n; i++)
        {
            if(i % 2 == 0)
            {
                even[a[i]] += 1;
            }
            else
            {
                odd[a[i]] += 1;
            }
        }
        
        
        
        int oddMax = 0;
        int evenMax = 0;
        int ans = 0;
        int oddMaxKey = 0;
        int evenMaxKey = 0;
        
        foreach(int key in odd.Keys)
        {
            oddMax = Math.Max(oddMax, odd[key]);
        }
        
        foreach(int key in even.Keys)
        {
            evenMax = Math.Max(evenMax, even[key]);
        }
        
        ans = (n / 2 - oddMax) + (n / 2 - evenMax);
        foreach(KeyValuePair<int, int> pair in even)
        {
            if(pair.Value == evenMax)
            {
                evenMaxKey = pair.Key;
            }
        }
        foreach(KeyValuePair<int, int> pair in odd)
        {
            if(pair.Value == oddMax)
            {
                oddMaxKey = pair.Key;
            }
        }
        if(evenMaxKey == oddMaxKey)
        {
            ans += Math.Min(evenMax, oddMax);
        }
        Console.WriteLine(ans);
    }
}