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>
using namespace std;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}


int nextInt() { int x; scanf("%d", &x); return x;}

int f(const vector<int> v) {
  int n = v.size();
  map<int, int> count;
  for (int x : v) count[x]++;
  int ma = 0;
  for (auto it = count.begin(); it != count.end(); ++it) {
    chmax(ma, it -> second);
  }
  return n - ma;
}

int main2() {
  int N = nextInt();

  vector<int> v0, v1;
  REP(i, N) {
    int x = nextInt();
    if (i % 2 == 0) v0.push_back(x);
    else v1.push_back(x);
  }
  sort(ALL(v0));
  sort(ALL(v1));
  if (v0[0] == v1[0] && v0[0] == v0.back()) {
    cout << v0.size() << endl;
    return 0;
  }
  int ans = f(v0) + f(v1);
  cout << ans << endl;
  return 0;
}

int main() {
  for (;!cin.eof();cin>>ws)
    main2();
  return 0;
}