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>

#define rep(i, n) for (int i = 0; i < (n); ++i)

using namespace std;

using ll = long long;
using P = pair<int, int>;

int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  rep(i, n) cin >> v[i];

  map<int, int> oddMp;
  map<int, int> evenMp;
  rep(i, n) {
    if (i % 2 == 0) {
      evenMp[v[i]]++;
    } else {
      oddMp[v[i]]++;
    }
  }
  vector<P> oddP;
  for (auto odd : oddMp) {
    oddP.emplace_back(odd.second, odd.first);
  }
  sort(oddP.rbegin(), oddP.rend());
  vector<P> evenP;
  for (auto even : evenMp) {
    evenP.emplace_back(even.second, even.first);
  }
  sort(evenP.rbegin(), evenP.rend());
  int res;
  if (oddP.at(0).second == evenP.at(0).second) {
    int v1 = n - oddP.at(0).first - (evenP.size() >= 2 ? evenP.at(1).first : (n / 2 - evenP.at(0).first));
    int v2 = n - (oddP.size() >= 2 ? oddP.at(1).first : (n / 2 - oddP.at(0).first)) - evenP.at(0).first;
    res = min(v1, v2);
  } else {
    res = n - (oddP.at(0).first + evenP.at(0).first);
  }
  cout << res << endl;

  return 0;
}