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)

/*
First, solve the problem.Then write the code. - John Johnson
@author: returnAC
*/
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define need_for_speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0); // Thanks Joshua :)
#define watch(x) cout<<"->"<<#x<<" : "<<x<<endl;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c)) 
typedef long long ll;
typedef unsigned long long llu;
const double PI = 2*acos(0.0);
const int MAXN = 100005;
int path1[MAXN];
int path2[MAXN];
signed main()
{
	//freopen("","r",stdin);
	//freopen("","w",stdout);
	need_for_speed;
	int n;
	cin>>n;
	vector<int> v(n);
	for(int i=0;i<n;i++)cin>>v[i];
	for(int i=0;i<n;i+=2)
		path1[v[i]]++;
	for(int i=1;i<n;i+=2)
		path2[v[i]]++;
	sort(path1,path1+MAXN); reverse(path1,path1+MAXN);
	sort(path2,path2+MAXN); reverse(path2,path2+MAXN);
	int res = 0;
	if(v[0]!=v[1])
		res = (n/2)-path1[0] + (n/2)-path2[0];
	else
		res = min( (n/2)-path1[0] + (n/2)-path2[1] , (n/2)-path1[1] + (n/2)-path2[0] );
	cout<<res<<endl;
}