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: Time Limit Exceeded)

#include<iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;
int Max = 100000;
typedef pair<int, int> count_s;

int main(){
  int n;
  cin >> n;
  vector<int> Even, Odd;
  for(int i=0;i<n;i++){
    int l;
    cin >> l;
    if(i%2==0) Even.push_back(l);
    else Odd.push_back(l);
  }

  vector<count_s> Even_s, Odd_s;
  int even_length = Even.size();
  int odd_length = Odd.size();

  for(int i=1;i<=Max;i++){
    int even_count = count(Even.begin(), Even.end(), i);
    Even_s.push_back(make_pair(even_length - even_count,i));

    int odd_count = count(Odd.begin(), Odd.end(), i);
    Odd_s.push_back(make_pair(odd_length-odd_count,i));
  }

  sort(Even_s.begin(),Even_s.end());
  sort(Odd_s.begin(),Odd_s.end());

  int ans;
  if(Even_s[0].second!=Odd_s[0].second){
  	ans = Even_s[0].first + Odd_s[0].first;
  	cout << ans <<endl;
  }else{
  	ans = min(Even_s[0].first + Odd_s[1].first, Even_s[1].first + Odd_s[0].first);
  	cout << ans <<endl;
  }

  return 0;
}