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 <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <random>

#define int long long
#define pb push_back
#define ff first
#define ss second
#define ld long double
#define rsz resize
#define ass assign

using namespace std;

mt19937 gen(time(nullptr));

const int INF = 1e6;

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> a(n), b(INF), c(INF);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (i % 2 == 0) {
            b[a[i]]++;
        } else c[a[i]]++;
    }
    for (int i = 0; i < INF; i++) {
        if (b[i] > 0 && c[i] > 0 && c[i] >= b[i]) b[i] = 0;
        if (b[i] > 0 && c[i] > 0 && b[i] >= c[i]) c[i] = 0;
    }
    int mx1 = -INF, mx2 = -INF;
    for (int i = 0; i < INF; i++) {
        mx1 = max(mx1, b[i]);
        mx2 = max(mx2, c[i]);
    }
    cout << n - mx1 - mx2;
    return 0;
}