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++ (GCC 9.2.1) (judge verdict: Wrong Answer)


#include<algorithm>     //sort,二分探索,など
#include<bitset>        //固定長bit集合
#include<cmath>         //pow,logなど
#include<complex>       //複素数
#include<deque>         //両端アクセスのキュー
#include<fstream>       //ファイルストリーム（標準入力変更用）
#include<functional>    //sortのgreater
#include<iomanip>       //setprecision(浮動小数点の出力の誤差)
#include<iostream>      //入出力
#include<iterator>      //集合演算(積集合,和集合,差集合など)
#include<map>           //map(辞書)
#include<numeric>       //iota(整数列の生成),gcdとlcm(c++17)
#include<queue>         //キュー
#include<set>           //集合
#include<stack>         //スタック
#include<string>        //文字列
#include<unordered_map> //イテレータあるけど順序保持しないmap
#include<unordered_set> //イテレータあるけど順序保持しないset
#include<utility>       //pair
#include<vector>        //可変長配列

//名前
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef map<string, int> msi;
typedef map<string, ll> msll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pllll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvll;
typedef vector<vector<string>> vvs;
typedef vector<vector<bool>> vvb;

//定数
const ll MOD = 1000000007;
const ll INF = 1000000000000000000;

//マクロ
#define rep(i,n) for(int i=0;i<n;i++)
#define reps(i,s,e) for(int i=s;i<e;i++)
#define repse(i,s,e) for(int i=s;i<=e;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define in1(x1) cin >> x1
#define in2(x1, x2) cin >> x1 >> x2
#define in3(x1, x2, x3) cin >> x1 >> x2 >> x3
#define inN(x, N) rep(i, N) in1(x[i])
#define outl(x) cout << x << endl
#define out2l(x, y) cout << x << " " << y << endl 

//よく使う関数
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

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

    //標準入力をファイルに変更
    //std::ifstream in("input.txt");
    //std::cin.rdbuf(in.rdbuf());
    int n;
    in1(n);

    vi v(n);
    inN(v, n);

    map<int, int> even, odd;

    rep(i, n)
    {
        if (i % 2 == 0) even[v[i]]++;
        else odd[v[i]]++;
    }

    if (odd.size() == 1 && even.size() == 1 && *odd.begin() == *even.begin())
    {
        outl(n / 2);
    }
    else
    {
        int ans = 0;

        int n11 = -1;
        int n12 = -1;
        int m11 = -1;
        int m12 = -1;

        for (auto x : even)
        {
            ans += x.second;
            if (m11 <= x.second)
            {
                n12 = n11;
                m12 = m11;
                n11 = x.first;
                m11 = x.second;
            }
        }

        int n21 = -1;
        int n22 = -1;
        int m21 = -1;
        int m22 = -1;

        for (auto x : odd)
        {
            ans += x.second;
            if (m21 <= x.second)
            {
                n22 = n21;
                m22 = m21;
                n21 = x.first;
                m21 = x.second;
            }
        }

        if (n11 != n21)
        {
            ans -= (m11 + m21);
        }
        else if (m12 != -1 && m22 != -1)
        {
            ans -= max(m11 + m22, m12 + m21);
        }
        else if (m12 != -1 && m22 == -1)
        {
            ans -= (m12 + m21);
        }
        else if (m12 == -1 && m22 != -1)
        {
            ans -= (m11 + m22);
        }

        outl(ans);
    }

    return 0;
}