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: Wrong Answer)

#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;

bool comp(pair<int,int>& a,pair<int,int>& b)
{
	return a.second>b.second;
}
int main()
{
	int n,a[50005],b[50005],p,q,ans;
	vector<pair<int,int> >ca,cb;
	cin>>n;
	for(int i=0;i<n/2;i++)cin>>a[i]>>b[i];
	sort(a,a+n/2);
	sort(b,b+n/2);
	p=q=1;
	ca.push_back(make_pair(0,0));
	cb.push_back(make_pair(0,0));
	ca.push_back(make_pair(a[0],1));
	cb.push_back(make_pair(b[0],1));
	for(int i=1;i<n/2;i++)
	{
		if(ca[p].first!=a[i])
		{
			ca.push_back(make_pair(a[i],1));
			p++;
		}
		else
		{
			ca[p].second++;
		}
		if(cb[q].first!=b[i])
		{
			cb.push_back(make_pair(b[i],1));
			q++;
		}
		else
		{
			cb[q].second++;
		}
	}
	sort(ca.begin(),ca.end(),comp);
	sort(cb.begin(),cb.end(),comp);
	p=q=0;
	if(ca[p].first==cb[q].first)
	{
		if(ca[p].second<cb[q].second)p++;
		else q++;
	}
	ans=n-ca[p].second-cb[q].second;
	cout<<ans<<endl;
	return 0;
}