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 <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;


int main(){
	int n;
	int x;
	cin>>n;
	map<int,int> mp1, mp2;
	for(int i=0;i<n;i++){
		cin>>x;
		if(i%2==0){
			mp1[x]++;
		}else{
			mp2[x]++;
		}
	}
	int max1=0;
	int max2=0;
	int max_v1=-1, max_v2=-2;
	int semi1=0;
	int semi2=0;
	for(auto itr = mp1.begin();itr != mp1.end();itr++){
		if(max1<itr->second){
			max_v1=itr->first;
		}
		max1 = max(max1, itr->second);
	}
	for(auto itr = mp1.begin();itr != mp1.end();itr++){
		
		if(itr->first!=max_v1){
			semi1 = max(semi1, itr->second);
		}
	}
	for(auto itr = mp2.begin();itr != mp2.end();itr++){
		if(max2<itr->second){
			max_v2=itr->first;
		}
		max2=max(max2, itr->second);
	}
	for(auto itr = mp2.begin();itr != mp2.end();itr++){
		
		if(itr->first!=max_v2){
			semi2 = max(semi2, itr->second);
		}
	}

//	cout<<max1<<" "<<max2<<endl;
	if(max_v1!=max_v2){
		cout<<n -max1-max2<<endl;
	}else {
		cout<<min(n-max1-semi2,n-max2-semi1)<<endl;
	}

	return 0;
}