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<iostream>

using namespace std;
int main()
{
	int N,i,v[50010],w[50010],cnt_even[100010],cnt_odd[100010];
	int max_even=-1,max_odd=-1,tmp_even,tmp_odd,emx=-1,omx=-1;
	cin >> N;
	for(i=0; i<=100000; i++){
		cnt_even[i] = 0;
		cnt_odd[i] = 0;
	}
	for(i=0; i<N/2; i++){
		cin >> v[i] >> w[i];
		cnt_even[v[i]]++;
		cnt_odd[w[i]]++;
	}
	for(i=0; i<=100000; i++){
		if(max_even < cnt_even[i]){
			max_even = cnt_even[i];
			tmp_even = i;
		}else if(emx < cnt_even[i]){
			emx = cnt_even[i];
		}
		if(max_odd < cnt_odd[i]){
			max_odd = cnt_odd[i];
			tmp_odd = i;
		}else if(omx < cnt_odd[i]){
			omx = cnt_odd[i];
		}
	}
	
	if(tmp_even != tmp_odd){
		cout << N - max_even - max_odd << endl;
	}else{
		if(max_even > max_odd){
			cout << N - max_even - omx << endl;
		}else if(max_even < max_odd){
			cout << N - emx - max_odd << endl;
		}else{
			if(emx < omx){
				cout << N - max_even - omx << endl;
			}else{
				cout << N - emx - max_odd << endl;
			}
		}
	}
	
	return 0;
}