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 <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define rep2(i, s, n) for(int i=s; i<(n); ++i)

using namespace std;

typedef long long ll;

static const int INTINF = (2147483647);
static const ll LLINF = (9223372036854775807);
static const int MAX = 100001;
static const ll MOD = 1000000007;

//--global--//

ll dp[101][101];

//----------//

ll cir(ll x, ll y, ll h, ll cx, ll cy){
    return h + abs(x-cx) + abs(y-cy);
}

int main(int argc, const char * argv[]) {
    //提出時、消す----//
    //--------------//
    
    int n;
    cin >> n;
    int v[MAX];
    int a[MAX];
    int b[MAX];
    int maxA = 0;
    int numA = 0;
    int maxB = 0;
    int numB = 0;
    int nexA = 0;
    int nexB = 0;
    priority_queue<pair<int, int> > q;
    rep(i, n){
        cin >> v[i];
        if(i%2==0){
            a[v[i]]++;
        }
        else{
            b[v[i]]++;
        }
    }

    rep(i, MAX){
        int num = i;
        int val = a[i];
        pair<int, int> f = make_pair(val, num);
        q.push(f);
    }

    maxA = q.top().first;
    numA = q.top().second;
    q.pop();
    nexA = q.top().first;
    while(!q.empty()){
        q.pop();
    }

    rep(i, MAX){
        int num = i;
        int val = b[i];
        pair<int, int> f = make_pair(val, num);
        q.push(f);
    }

    maxB = q.top().first;
    numB = q.top().second;
    q.pop();
    nexB = q.top().first;
    while(!q.empty()){
        q.pop();
    }


    cout << maxA << ", " << numA << ", " << nexA << endl;
    cout << maxB << ", " << numB << ", " << nexB << endl;


    int res = 0;
    if(numA == numB){
        if(nexA >= nexB){
            res = n-nexA - maxB;
        }
        else{
            res = n-nexB - maxA;
        }
    }
    else{
        res = n-maxA - maxB;
    }

    cout << res << endl;


    
    return 0;
}