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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int main() {
  int n;
  cin >> n;
  map<int, int> odd;
  map<int, int> even;
  rep(i, n) {
    int a;
    cin >> a;
    if ((i + 1) % 2 == 0) {
      even[a]++;
    } else {
      odd[a]++;
    }
  }

  vector<pair<int, int>> odd2;
  for (auto kv : odd) {
    odd2.push_back(make_pair(kv.second, kv.first));
  }
  sort(odd2.rbegin(), odd2.rend());

  vector<pair<int, int>> even2;
  for (auto kv : even) {
    even2.push_back(make_pair(kv.second, kv.first));
  }
  sort(even2.rbegin(), even2.rend());

  if (odd2[0].second != even2[0].second) {
    cout << n - odd2[0].first - even2[0].first << endl;
    return 0;
  }

  if (odd2[0].first == n / 2) {
    cout << n / 2 << endl;
    return 0;
  }

  if (odd2[0].first > even2[0].first) {
    cout << n - odd2[0].first - even2[1].first << endl;
  } else {
    cout << n - odd2[1].first - even2[0].first << endl;
  }
  return 0;
}