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>

using namespace std;

int n,v[100005],cnt[100005],maks,temp1,temp2,num1,num2,temp3,mboh2,mboh1;

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>v[i];
	}
	memset(cnt,0,sizeof(cnt));
	maks=0;
	for(int i=0;i<n;i=i+2)
	{
		cnt[v[i]]++;
	}
	for(int i=1;i<=1e5;i++)
	{
		maks=max(maks,cnt[i]);
		if(maks==cnt[i]) num1=i;
	}
	temp1=(n/2)-maks;
	memset(cnt,0,sizeof(cnt));
	maks=0;
	for(int i=1;i<n;i=i+2)
	{
		cnt[v[i]]++;
	}
	for(int i=1;i<=1e5;i++)
	{
		maks=max(maks,cnt[i]);
		if(maks==cnt[i]) num2=i;
	}
	temp2=(n/2)-maks;
	//cout<<num1<<" "<<num2<<endl;
	if(num1!=num2) cout<<temp1+temp2<<endl;
	else
	{
		maks=0;
		cnt[num2]=0;
		for(int i=1;i<=1e5;i++)
		{
			maks=max(maks,cnt[i]);
		}
		temp3=(n/2)-maks;
		if(temp3>temp1)
		{
			maks=0;
			memset(cnt,0,sizeof(cnt));
			for(int i=0;i<n;i=i+2)
			{
				cnt[v[i]]++;
			}
			cnt[num1]=0;
			for(int i=1;i<=1e5;i++)
			{
				maks=max(maks,cnt[i]);
			}
			int temp4=(n/2)-maks;
			//cout<<temp4<<" "<<temp2<<endl;
			cout<<temp4+temp2<<endl;
		}
		else
		{
			//cout<<temp1<<" "<<temp2<<endl;
			cout<<temp1+temp3<<endl;
		}
	}
}