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 rep(i,N) for(int i=0;i<int(N);++i)
#define rep1(i,N) for(int i=1;i<int(N);++i)
#define all(a) (a).begin(),(a).end()		//sort(all(vi S)) sort(all(string S))
#define print(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<_<<", "; cerr<<"]"<<endl; }
#define printpair(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<"{"<<_.first<<","<<_.second<<"}"<<", "; cerr<<"]"<<endl; }

using P = pair<int, int>;		//P.first, P.second
typedef long long ll;
typedef vector<int> vi;
typedef set<int> seti;
typedef vector<string> vs;

const int MOD = 1e9+7;
const int INF = 1e9;

int main() {
	int N;
	cin>>N;
	vector<int> v1,v2;
	std::map<ll, ll> map1,map2;
	rep(i,N){
		int v;
		cin>>v;
		if(i%2 == 0){
			v1.push_back(v);
			map1[v]++;
		}
		else{
			v2.push_back(v);
			map2[v]++;
		}
	}
	vector<P> cnt1,cnt2;
	//個数、keyのPairのベクター
	for(auto x :  map1){
		cnt1.push_back({x.second,  x.first});
	}
	for(auto x :  map2){
		cnt2.push_back({x.second,  x.first});
	}
	cnt1.push_back({0,0});
	cnt2.push_back({0,0});
	sort(all(cnt1));
	sort(all(cnt2));
	printpair(cnt1);printpair(cnt2);
	int ans  = N;
	int N1 = cnt1.size();
	int N2 = cnt2.size();
	if(cnt1[N1-1].second==cnt2[N2-1].second){
		ans -= max(cnt1[N1-1].first+cnt2[N2-2].first,cnt1[N1-2].first+cnt2[N2-1].first);
	}
	else{
		ans -= cnt1[N1-1].first+cnt2[N2-1].first;
	}
	cout<<ans<<endl;
}