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 <sstream>
#include <cstdio>
#include <cmath>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <cstring>

#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))

#define INF (int)1e9+7

static const int dy[4] = {0,1,0,-1};
static const int dx[4] = {1,0,-1,0};

typedef long long ll;

using namespace std;

map<int, int> am, bm;

int main(){
    int n;

    cin >> n;

    rep(i, n){
        int in;
        cin >> in;
        if(i % 2 == 0) am[in]++;
        else bm[in]++;
    }
    int max1 = n;
    int max2 = n;
    int a = 0, b = 0;

    for(auto e : am){
        if(n/2 - e.second < max1){
            max1 = min(max1, n/2 - e.second);
            a = e.first;
        }
    }
    for(auto e : bm){
        if(n/2 - e.second < max2){
            max2 = min(max2, n/2 - e.second);
            b = e.first;
        }
    }
    if(a != b){
        cout << max1 + max2 << endl;
        return 0;
    }

    int ans = n;
    for(auto e : am){
        if(e.first != a) ans = min(ans, max2 + n/2 - e.second);
    }
    //if(am.size() == 1) ans = min(ans, max2 + n/2);

    for(auto e : bm){
        if(e.first != b) ans = min(ans, max1 + n/2 - e.second);
    }
    //if(bm.size() == 1) ans = min(ans, max1 + n/2);
        
    cout << ans << endl;

    return 0;
}