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 <bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;

int main(void) {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<int> even(n / 2, 0);
  vector<int> odd(n / 2, 0);
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      even[i / 2] = v[i];
    } else {
      odd[i / 2] = v[i];
    }
  }
  sort(even.begin(), even.end());
  sort(odd.begin(), odd.end());

  even.push_back(0);
  odd.push_back(0);

  P even_min = P(0, 0);
  P even_second = P(0, 0);
  P odd_min = P(0, 0);
  P odd_second = P(0, 0);

  for (int i = 0; i < n / 2; i++) {
    if (even[i] != even[i + 1]) {
      even_min = P(even[i], i + 1);
      break;
    }
  }
  for (int i = even_min.second; i < n / 2; i++) {
    if (even[i] != even[i + 1]) {
      even_second = P(even[i], i + 1 - even_min.second);
      break;
    }
  }
  for (int i = 0; i < n / 2; i++) {
    if (odd[i] != odd[i + 1]) {
      odd_min = P(odd[i], i + 1);
      break;
    }
  }
  for (int i = odd_min.second; i < n / 2; i++) {
    if (odd[i] != odd[i + 1]) {
      odd_second = P(odd[i], i + 1 - odd_min.second);
      break;
    }
  }

  int ans;
  cout << even_min.second << " " << even_second.second << " " << odd_min.second
       << " " << odd_second.second << endl;
  if (even_min.first == odd_min.first) {
    ans = min(n / 2 - even_min.second + n / 2 - odd_second.second,
              n / 2 - even_second.second + n / 2 - odd_min.second);
  } else {
    ans = n / 2 - even_min.second + n / 2 - odd_min.second;
  }

  cout << ans << endl;
  return 0;
}