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 (ll i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
using P = pair<int,int>;
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cout << H << " "; debug_out(T...); }
#ifdef _DEBUG
  #define debug(...) debug_out(__VA_ARGS__)
#else
  #define debug(...) 
#endif

int main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(15);
  int n;
  cin >> n;
  vector<P> cntodd(100005), cnteve(100005);
  rep(i,100005) {
    cntodd[i].second = i;
    cnteve[i].second = i;
  }
  rep(i,n) {
    int a;
    cin >> a;
    if (i % 2 == 0) cnteve[a].first++;
    else cntodd[a].first++;
  } 
  sort(all(cnteve), greater<P>());
  sort(all(cntodd), greater<P>());
  int ans = 0;
  auto dif = [&](P p) {
    return fabs(p.first - n/2);
  };
  if (cnteve[0].second == cntodd[0].second) { //数が同じ
    ans = min(dif(cnteve[0]) + dif(cntodd[1]), dif(cnteve[1]) + dif(cntodd[0]));
    cout << ans << endl;
    return 0;
  }
  ans = dif(cnteve[0]) + dif(cntodd[0]);
  cout << ans << endl;
  return 0;
}