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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <bits/stdc++.h>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <map>

using namespace std;

int main(void)
{
    long long n;
    cin >> n;
    map<long long, long long> mpodd;
    map<long long, long long> mpeven;

    for (int i = 1; i <= n; i++)
    {
        long long tmp;
        cin >> tmp;
        if (i % 2 == 1)
        {
            mpodd[tmp] += 1;
        }
        else
        {
            mpeven[tmp] += 1;
        }
    }

    long long countodd = 0;
    long long maxodd = -1;
    for (auto i : mpodd)
    {
        countodd = max(countodd, i.second);
        maxodd = i.first;
    }

    long long counteven = 0;
    long long maxeven = -1;
    for (auto i : mpeven)
    {
        counteven = max(counteven, i.second);
        maxeven = i.first;
    }

    if (maxodd != maxeven)
    {
        cout << n - countodd - counteven << endl;
    }
    else
    {
        mpodd[maxodd] = 0;
        mpeven[maxeven] = 0;

        long long countodd2 = -1, counteven2 = -1;
        for (auto i : mpodd)
        {
            countodd2 = max(countodd2, i.second);
        }

        for (auto i : mpeven)
        {
            counteven2 = max(counteven2, i.second);
        }

        if (countodd2 == -1)
            countodd2 = n / 2;
        if (counteven2 == -1)
            counteven2 = n / 2;

        long long sum;
        sum = max(countodd + counteven2, countodd2 + counteven);
        cout << n - sum << endl;
    }

    return 0;
}