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)

//************************
//*  Date: 2018.9.22     *
//*  Problem:            *
//*  Author: xuziyuan    * 
//************************
 
#include <bits/stdc++.h>
 
#define rt0 return 0
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define replet(c) for(char c='a';c<='z';c++)
#define LL long long
#define pii pair<int,int>
#define pb push_back
#define fi first
#define se second
#define mpr make_pair
#define sqr(a) ((a)*(a))
 
using namespace std;
 
const LL MOD=1e9+7;
 
int n,a[100010];
int ev[100010],od[100010];
 
int main()
{
	cin>>n;
	repn(i,n)
	{
		cin>>a[i];
		if(i%2==1) od[a[i]]++;
		else ev[a[i]]++;
	}
	int mxe=0,mxo=0,mxe2=0,mxo2=0;
	int mxec,mxoc,mxec2,mxoc2;
	rep(i,100005)
	{
		if(ev[i]>=mxe) mxe=ev[i],mxec=i;
		else if(ev[i]>=mxe2) mxe2=ev[i],mxec2=i;
		if(od[i]>=mxo) mxo=od[i],mxoc=i;
		else if(od[i]>=mxo2) mxo2=od[i],mxoc2=i;
	}
	if(mxec==mxoc)
	{
		cout<<min(n-mxe-mxo2,n-mxo-mxe2)<<endl;
		rt0;
	}
	cout<<n-mxe-mxo<<endl;
	rt0;
}