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 <bits/stdc++.h>
#define F first
#define S second
#define PB push_back
#define REP(i,a,b) for (int i = a ; i <= b ; i++)
#define REPI(i,a,b) for (int i = a ; i >= b ; i--)
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;

const int OO = 1e9+7;
const int MOD = 1e9+7;
const int N = 1e5+7;

int cnt1[N];
int cnt2[N];

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    //freopen("teleport.in","r",stdin);
    //freopen("teleport.out","w",stdout);
	
	int n;
	scanf("%d",&n);
	REP(i,0,n-1)
	{
		int x;
		scanf("%d",&x);
		if(i&1) cnt1[x]++;
		else cnt2[x]++;
	}
	int mx1,mx2;
	int mx11,mx22;
	mx1=mx2=0;
	mx11=mx22=0;
	REP(i,0,N-1)
	{
		if (cnt1[i] > cnt1[mx1])
		{
			mx11 = mx1;
			mx1 = i;
		}
		else if (cnt1[i] > cnt1[mx11]) mx11 = i;

		if (cnt2[i] > cnt1[mx2])
                {
                        mx22 = mx2;
                        mx2 = i;
                }
                else if (cnt2[i] > cnt2[mx22]) mx22 = i;
	}
	if (mx1 != mx2)
	{
		printf("%d",n-(cnt1[mx1]+cnt2[mx2]));
	}
	else
	{
		int ans = max(cnt1[mx11]+cnt2[mx2],cnt1[mx1]+cnt2[mx22]);
		printf("%d",n-ans);
	}
    return 0;
}