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++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
ll GCD(ll a,ll b){return b?GCD(b,a%b):a;}
ll LCM(ll a,ll b){return (a*b)/GCD(a,b);}
int abs(int x){return x>0?x:-x;}
 
//const int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
//const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

int main()
{
	int n;
	cin >> n;
	int a[n];
	map<int , int> m;
	REP(i, n){
		cin >> a[i];
		m[a[i]]++;
	}

	int max1 = 0, max2 = 0;
	
	if ((int)m.size() == 1){
		cout << n / 2 << endl;
		return 0;
	}
	
	int key;
	for (auto x : m){
		if (max1 < x.second){
			max1 = x.second;
			key = x.first;
		}
	}

	m.erase(key);

	for (auto x : m){
		if (max2 < x.second){
			max2 = x.second;
		}
	}

	cout << n - (max1 + max2) << endl;
}