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 <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
const ll mod = 1000000007;

int main() {
    int n;
    cin >> n;
    int m = n/2;

    int v[n];
    rep(i, n) {
        cin >> v[i];
    }

    int e[m], o[m];
    for(int i = 0; i < n; i+= 2) {
        e[i] = v[i];
    }
    for(int i = 1; i < n; i+= 2) {
        o[i] = v[i];
    }

    sort(e, e+m); 
    int te = e[0];
    int ke = 0;
    int ce[m];   //e[m]の小さい数から順にそれぞれ何個ずつ含まれているかをce[m]に入れる
    rep(i, m) {
        ce[i] = 0;
    }   
    rep(i, m) {
        if(te < e[i]) {
            te = e[i];
            ke++;
            ce[ke]++;
        } 
        else {
            ce[ke]++;
        }
    }

    sort(o, o+m); 
    int to = o[0];
    int ko = 0;
    int co[m];   //o[m]の小さい数から順にそれぞれ何個ずつ含まれているかをco[m]に入れる
    rep(i, m) {
        co[i] = 0;
    }   
    rep(i, m) {
        if(to < o[i]) {
            to = o[i];
            ko++;
            co[ko]++;
        } 
        else {
            co[ko]++;
        }
    }

    sort(ce, ce+m);
    reverse(ce, ce+m);
    sort(co, co+m);
    reverse(co, co+m);

    int ans = n - (ce[0] + co[0]);

    cout << ans << endl;
    return 0;
}