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 rrep(i,n) for (int i = n-1; i >= 0; i--)
#define rep2(i,s,n) for (int i = s; i < n; ++i)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define tmax(a,b,c) max(a,max(b,c))
#define tmin(a,b,c) min(a,min(b,c))
#define pb push_back
using namespace std;
using ll = long long;
using P = pair<int,int>;
using LP = pair<ll,ll>;
template<class T> inline bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> inline bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
const int inf = 1001001001;
const ll linf = 1001001001001001001;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<int> v(n);
    map<int,int> a,b;
    rep(i,n) {
        if(i%2 == 0) a[v[i]]++;
        else b[v[i]]++;
    }
    int amx=0,bmx=0;
    int na,nb;
    int ia,ib;
    for(auto p : a) {
        if(p.second > amx) {
            na = 1;
            amx = p.second;
            ia = p.first;
        } else if(p.second == amx) {
            na++;
        }
    }
    for(auto p : b) {
        if(p.second > bmx) {
            nb = 1;
            bmx = p.second;
            ib = p.first;
        } else if(p.second == bmx) {
            nb++;
        }
    }
    if(na >= 2 || nb >= 2) cout << n-amx-bmx << endl;
    else if(ia != ib) cout << n-amx-bmx << endl;
    else {
        int namx = 0, nbmx = 0;
        for (auto p : a) {
            if (p.second > namx && p.second < amx) {
                namx = p.second;
            }
        }
        for (auto p : b) {
            if (p.second > nbmx && p.second < bmx) {
                nbmx = p.second;
            }
        }
        cout << min(n-amx-nbmx,n-bmx-namx) << endl;
    }   
}