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>
	using namespace std;
	#define IOS ios::sync_with_stdio(false), cin.tie(0)
	#define REP(i, n) for (int i = 0; i < n; i++)
	#define FOR(i, a, b) for (int i = a; i < b; i++)
	#define ALL(x) (x).begin(), (x).end()
	#define dump(x) cout << (x) << endl
	#define LMAX 9223372036854775807LL
	#define LMIN -9223372036854775807LL

	using ll = long long;
	template<typename T>
	using vec = vector<T>;
	using P = pair<ll, ll>;

	template <typename A, size_t N, typename T>
	void Fill(A (&array)[N], const T &val){
		fill((T *)array, (T *)(array + N), val);
	}
	
	const ll INF = 1e18;
	const ll mod = 1e9 + 7;

	//int dx[4] = {1,0,-1,0};
	//int dy[4] = {0,1,0,-1};
	//ll pas[51][52];
	
	signed main(){
		IOS;
		ll n;
		cin >> n;
		map<ll,ll> mp1,mp2;
		REP(i,n){
			int x;
			cin >> x;
			if(i % 2)mp1[x]++;
			else mp2[x]++;
		}
		vec<P> p1,p2;
		for(auto it = mp1.begin();it != mp1.end();it++){
			ll a = it->first;
			ll b = it->second;
			p1.push_back(P(b,a));
		}
		for(auto it = mp2.begin();it != mp2.end();it++){
			ll a = it->first;
			ll b = it->second;
			p2.push_back(P(b,a));
		}
		sort(ALL(p1),greater<P>());
		sort(ALL(p2),greater<P>());
		ll ans = 0;
		int s = p1.size();
		int t = p2.size();
		for(int i = 0;i < min(2, s);i++){
			for(int j = 0;j < min(2, t);j++){
				if(p1[i].second == p2[j].second)continue;
				ans = max(ans,p1[i].first + p2[j].first);
			}
		}
		if(ans == 0){
			dump(n / 2);
			return 0;
		}
		dump(n - ans);
	}