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>
#include <algorithm>
#include <numeric>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <unordered_set>
#include <unordered_map>
#include <cmath>
using namespace std;
typedef long long ll;
typedef unordered_set<ll> uset;
typedef unordered_map<ll,ll> umap;
typedef pair<ll,ll> pr;

void print(const vector<pr>& v ){
  for( auto p : v ) cerr << " (" << p.first << "," << p.second << ")";
  cerr << endl;
}
 
void print( const vector<int>& v ){
  for( auto x : v ) cerr << " " << x; cerr << endl;
}
 
int main( int argc, char* argv[] ){
  int N;
  cin >> N;
  int N2 = N/2;
  vector<ll> v1(N2), v2(N2);
  for( int i=0; i<N2; ++i ){
      cin >> v1[i];
      cin >> v2[i];
  }
  umap c1, c2;
  for( auto x : v1){
      c1[x]++;
  }
  for( auto x : v2){
      c2[x]++;
  }

  vector<pr> l1 = vector<pr>( c1.begin(), c1.end() );
  vector<pr> l2 = vector<pr>( c2.begin(), c2.end() );
  // print(l1);
  // print(l2);
  std::sort( l1.begin(), l1.end(), [](const pr& p1, const pr& p2) { return p1.second > p2.second; });
  std::sort( l2.begin(), l2.end(), [](const pr& p1, const pr& p2) { return p1.second > p2.second; });
  // print(l1);
  // print(l2);
  if( l1[0].first != l2[0].first ){
      cout << N - l1[0].second - l2[0].second << endl;
      return 0;
  }
  if( l1.size() == 1 && l2.size() == 1 ){
      cout << N2 << endl;
      return 0;
  }
  if( l1.size() == 1 ){
      //cout << N - l1[0].second - l2[1].second << endl;
      cout << 0 << endl;
      return 0;
  }
  if( l2.size() == 1 ){
      //cout << N - l1[1].second - l2[0].second << endl;
      cout << 0 << endl;
      return 0;
  }
  auto ret = std::max( N-l1[0].second-l2[1].second, N-l1[1].second-l2[0].second );
  cout << ret << endl;
  return 0;
}