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 ans, n, v[(int)1e5];
int odd[(int)1e5+1], even[(int)1e5+1];

int main(){
	cin >> n;
	for(int i = 0; i < n; ++i){
		cin >> v[i];
		if(i % 2 == 0) odd[v[i]]++;
		else even[v[i]]++;
	}
	
	int odd_first = 0, odd_first_itr = 0, odd_second = 0, odd_second_itr = 0;
	int even_first = 0, even_first_itr = 0, even_second = 0, even_second_itr = 0;

	for(int i = 0; i <= (int)1e5; ++i){
		if(odd[i] > odd_first){
			odd_first = odd[i];
			odd_first_itr = i;
		}
	}
	for(int i = 0; i <= (int)1e5; ++i){
		if(i == odd_first_itr) continue;
		if(odd[i] > odd_second){
			odd_second = odd[i];
			odd_second_itr = i;
		}
	}
	for(int i = 0; i <= (int)1e5; ++i){
		if(even[i] > even_first){
			even_first = even[i];
			even_first_itr = i;
		}
	}
	for(int i = 0; i <= (int)1e5; ++i){
		if(i == even_first_itr) continue;
		if(even[i] > even_second){
			even_second = odd[i];
			even_second_itr = i;
		}
	}
	
	if(odd_first_itr == even_first_itr){
		ans = min(n - odd_first - even_second, n - odd_second - even_first);
	}else{
		ans = n - odd_first - even_first;
	}

	cout << ans << endl;
}