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 <stdio.h>
#include <string>
#include <iostream>
#include <iomanip>
#include <map>
#include <algorithm>
#include <vector>
#include <math.h>
#define INF 1000000007
using namespace std;

int main(void){
	int n;
	cin >> n;
	vector<int> a(n/2),b(n/2);
	for(int i=0; i<n/2; i++)cin >> a[i] >> b[i];
	int odd_max=0,even_max=0,odd_key=a[0],even_key=b[0];
	int count=1,odd_second=0,even_second=0;
	sort(a.begin(),a.end());
	sort(b.begin(),b.end());
	for(int i=1; i<n/2; i++){
		if(a[i-1]==a[i])count++;
		else {
			if(count>odd_max){
				odd_second=odd_max;
				odd_max=count;
				odd_key=a[i-1];
			}else odd_second=max(odd_second,count);
			count=1;
		}
	}
	if(count>odd_max){
		odd_second=odd_max;
		odd_max=count;
		odd_key=a[n/2-1];
	}else odd_second=max(odd_second,count);
	if(odd_max==0)odd_max=n/2;
	count=1;
	for(int i=1; i<n/2; i++){
		if(b[i-1]==b[i])count++;
		else {
			if(count>even_max){
				even_second=even_max;
				even_max=count;
				even_key=b[i-1];
			}else even_second=max(even_second,count);
			count=1;
		}
	}
	if(count>even_max){
		even_second=even_max;
		even_max=count;
		even_key=b[n/2-1];
	}else odd_second=max(odd_second,count);
	if(even_max==0)even_max=n/2;
	int ans=0;
	if(even_key==odd_key)ans=min(n-odd_max-even_second,n-even_max-odd_second);
	else ans=n-odd_max-even_max;
	cout << ans << endl;
	return 0;
}