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 counta[100001],countb[100001];
int maxia,maxib,maxic,maxid;
int kea,keb,kec,ked;

int main() {
	int n;
	cin>>n;
	int a[n+1],b[n+1];
	for(int i=1; i<=n/2; i++) {
		cin>>a[i]>>b[i];
	}
	
	for(int i=1; i<=n/2; i++) {
		counta[a[i]]++;
	}
	for(int i=1; i<=n/2; i++) {
		countb[b[i]]++;
	}
	priority_queue<int> haha,hbhb;
	for(int i=1; i<=100000; i++) {
		if(counta[i]>maxia) {
			maxia=counta[i];
			kea=i;
		}
		if(countb[i]>maxib) {
			maxib=countb[i];
			keb=i;
		}
		haha.push(counta[i]);
		hbhb.push(countb[i]);
	}
	maxia=haha.top();
	haha.pop();
	maxid=haha.top();
	haha.pop();
	maxib=hbhb.top();
	hbhb.pop();
	maxid=hbhb.top();
	hbhb.pop();
	int hasil=0;
	if(kea==keb) {
		if(maxia+maxic>maxib+maxid) {
			hasil+=n-maxia-maxic;
		}else {
			hasil+=n-maxib-maxid;
		}
	}else {
		hasil+=(n/2)-maxia;
		hasil+=(n/2)-maxib;
	}
	cout<<hasil<<endl;
}