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;
using ll = long long;

const long long inf = 1e9;
#define fr(i,a,n) for (int i=a;i<n;i++)
#define rfr(i,n,a) for (int i=n;i>=a;i--)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll> 
#define endl "\n"
typedef vector<int> VI;
typedef vector<ll> VL;
typedef priority_queue<int> Pqi;
typedef priority_queue<ll> Pql;

int n, m, k;

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n;
	VI v(n+1);
	fr(i,0,n) {
		cin >> v[i+1];
	}
	map<int,int> ma1,ma2;
	for (int i = 1;i <= n-1;i+= 2) {
		ma1[v[i]]++;
		ma2[v[i+1]]++;
	}
	vector<pii> v1,v2;
	for (auto x:ma1) {
		v1.pb({(n/2) - x.second,x.first});
	}
	for (auto x:ma2) {
		v2.pb({(n/2) - x.second,x.first});
	}
	sort(v1.begin(),v1.end());
	sort(v2.begin(),v2.end());
	if (v1[0].second != v2[0].second) {
		cout << v1[0].first + v2[0].first << endl;
		return 0;
	}
	int ans = 1e9;
	if (v1.size() == 1 && v2.size() == 1) {
		cout << min(v1[0].first,v2[0].first) + (n/2) << endl;
		return 0;
	}
	if (v1.size() == 1) {
		ans = min(v1[0].first + v2[1].second,v2[0].first + (n/2));
		cout << ans << endl;
		return 0;
	}
	if (v2.size() == 1) {
		ans = min(v2[0].first + v1[1].second,v1[0].first + (n/2));
		cout << ans << endl;
		return 0;
	}
	ans = min(v2[0].first + v1[1].first,v1[0].first + v2[1].first);
	cout << ans << endl;
 	return 0;
}