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 <math.h>
//#include <numeric>
#include <vector>
#include <map>
#include <algorithm>
#include <queue>
#include <tuple>

using namespace std;
#define n_max 100000

#define rep(i,n) for(int i = 0; i < n; i++)
#define rep2(i,a,n) for(int i = a; i < n; i++)
#define in(a) cin >> a
#define out(a) cout << a << endl
typedef long long ll;
typedef pair<ll, ll> PI;

int main() {
    int n; cin >> n;
    vector<int> v(n);
    vector<tuple<int, int, int> > V;

    rep(i,n) { cin >> v[i]; }
    
    int index = 0;
    while (index < n-2) {
        bool even = true, odd = true;
        int cnt = 0;
        int j = index;
        while((even & odd) && (j < n-2)) {
            
            if(v[j] != v[j+2]) { even = false; }  
            else               { even = true;  }

            if(v[j+1] != v[j+3]) { odd = false; }  
            else                 { odd = true;  }
            j+=2;
            cnt++;
        }
        V.push_back(make_tuple(cnt, v[index], v[index+1]));
        index += 2 * (cnt + 1);
    }
    V.push_back(make_tuple(0, v[n-2], v[n-1]));

    int c_max = -1;
    int v_i1 = 0, v_i2 = 0; 
    for(int i = 0; i < V.size(); i++) {
        if(get<0>(V[i]) >= c_max) {

            if(v_i1 == get<1>(V[i]) && v_i2 == get<2>(V[i])) {
                c_max += get<0>(V[i]);
            } else {
                c_max = get<0>(V[i]);
                v_i1 = get<1>(V[i]);
                v_i2 = get<2>(V[i]);
            }
            //cout << get<1>(V[i]) << "," << get<2>(V[i]) << endl;
        }
    }
    int ans = 0;
    for(int i = 0; i < V.size(); i++) {
        if(v_i1 != get<1>(V[i])) { ans++; }
        if(v_i2 != get<2>(V[i])) { ans++; }
    }
    cout << ans << endl;
    return 0;
}