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>

using namespace std;

int main()
{
  int s_count[100001] {};
  int t_count[100001] {};
  int n;
  cin >> n;

  for (auto i = 0; i < n; i++) {
    int v;
    cin >> v;
    if (i % 2 == 0) {
      s_count[v]++;
    } else {
      t_count[v]++;
    }
  }

  int s_max1 = s_count[1];
  int s_max2 = 0;
  int t_max1 = t_count[1];
  int t_max2 = 0;
  int s_max_v = 1;
  int t_max_v = 1;

  for (auto i = 2; i <= 100000; i++) {
    if (s_max1 <= s_count[i]) {
      s_max_v = i;
      s_max2 = s_max1;
      s_max1 = s_count[i];
    }
    if (t_max1 <= t_count[i]) {
      t_max_v = i;
      t_max2 = t_max1;
      t_max1 = t_count[i];
    }
  }

  int ans = n;
  if (s_max_v != t_max_v) {
    ans -= s_max1 + t_max1;
  } else {
    if (s_max1 + t_max2 < t_max1 + s_max2) {
      ans -= t_max1 + s_max2;
    } else {
      ans -= s_max1 + t_max2;
    }
  }
  cout << ans << endl;
}