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: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
using P = pair<int, int>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n; cin >> n;
    map<int, int> m1, m2;
    for (int i = 0; i < n; i+=2) {
        int v1, v2; cin >> v1 >> v2;
        m1[v1]++;
        m2[v2]++;
    }

    vector<P> a1, a2;
    a1.push_back(make_pair(0, 0));
    a2.push_back(make_pair(0, 0));
    for (P p : m1) a1.push_back(p);
    for (P p : m2) a2.push_back(p);
    sort(a1.begin(), a1.end(), [](P p1, P p2) {
        return p1.second > p2.second;
    });
    sort(a2.begin(), a2.end(), [](P p1, P p2) {
        return p1.second > p2.second;
    });

    int ans;
    if (a1[0].first != a2[0].first)
        ans = n - (a1[0].second + a2[0].second);
    else
        ans = n - max(a1[0].second + a2[1].second,
                      a1[1].second + a2[0].second);
    cout << ans << "\n";
}