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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main()
{
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++)
  {
    cin >> v.at(i);
  }

  map<int, int> even, odd; //mapで各数列の要素の出現数を記録する

  for (int i = 0; i < n; i++)
  {
    if (i % 2 == 0)
      odd[v[i]]++; // キーvi(first)のときの値(second)をインクリメント
    else
      even[v[i]]++; // キーviのときの値をインクリメント
  }

  int even_1 = 0, even_2 = 0, odd_1 = 0, odd_2 = 0;
  int e1 = 0, e2 = 0, o1 = 0, o2 = 0;
  int cnt = 0;
  for (auto itr = even.begin(); itr != even.end(); ++itr)
  {
    if (itr->second > cnt)
    {
      cnt = itr->second;
      even_1 = itr->first;
    }
  }
  e1 = cnt;
  cnt = 0;
  for (auto itr = even.begin(); itr != even.end(); ++itr)
  {
    if (itr->second > cnt && itr->first != even_1) // 最も出現回数が大きいeven_1を除いた要素の中で最も出現回数が大きいものを求めている．
    {
      cnt = itr->second;
      even_2 = itr->first;
    }
  }
  e2 = cnt;

  cnt = 0;
  for (auto itr = odd.begin(); itr != odd.end(); ++itr)
  {
    if (itr->second > cnt)
    {
      cnt = itr->second;
      odd_1 = itr->first;
    }
  }
  o1 = cnt;
  cnt = 0;
  for (auto itr = odd.begin(); itr != odd.end(); ++itr)
  {
    if (itr->second > cnt && itr->first != odd_1)
    {
      cnt = itr->second;
      odd_2 = itr->first;
    }
  }
  o2 = cnt;

  if (even_1 == odd_1)
  {
    cout << min(n - e1 - o2, n - e2 - o1) << endl;
  }
  else
  {
    cout << n - e1 - o1 << endl;
  }

  return 0;
}