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>
#include "bits/stdc++.h"
using namespace std;
#define print(x) cout<<x<<endl;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a) for(int i=0;i<a;i++)
#define debug int debug = 0;
#define RREP(i,a) for(int i=a-1;i>=0;i--)
#define rrep(i,a,b) for(int i=a-1;i>=b;i--)
#define printall(n,array) {for(int i=0;i<n;i++){cout<<array[i]<<" ";}cout<<endl;}
#define U() cout<<endl;
typedef long long ll;
typedef pair<int, int> Pi;
typedef pair<int, Pi> V;
typedef vector<int> VE;
const ll mod = 1e9 + 7; //10^9+7

int main(){
    int Odd[100002];
    int Even[100002];
    int n;
    int q;
    int max_odd = 0;
    int max_even = 0;
    int max_odd_i = 0;
    int max_even_i = 0;
    int semi_max_odd = 0;
    int semi_max_even = 0;

    cin >> n;
    REP(i, n){
        cin >> q;
        if(i % 2 == 1){
            Odd[q]++;
            if(Odd[q] >= max_odd){
                if(max_odd_i != q)semi_max_odd = max_odd;
                max_odd = Odd[q];
                max_odd_i = q;
            }
        }else{
            Even[q]++;
            if(Even[q] >= max_even){
                if(max_even_i != q)semi_max_even = max_even;
                max_even = Even[q];
                max_even_i = q;
            }
        }
    }
    int cnt = 0;
    /*
    6
    5 3 5 2 5 5
    */
    if(max_odd_i == max_even_i){
        if(max_odd == max_even && max_odd == n / 2){
            print(n / 2);
            return 0;
        }
        if(semi_max_odd > semi_max_even){
            cnt += n / 2 - semi_max_odd;
            cnt += n / 2 - max_even;
        }else{
            cnt += n / 2 - semi_max_even;
            cnt += n / 2 - max_odd;
        }
        print(cnt);
        return 0;
    }else{
        cnt = cnt + n / 2 - max_odd;
        cnt = cnt + n / 2 - max_even;
        print(cnt);
    }
}