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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1e9+7;
int main() {
    ll n;
    cin >> n;
    vector<ll>a(n);
    vector<ll>odd;
    vector<ll>even;
    for(ll i=0;i<n;i++){
        cin >> a[i];
        if(i%2==0) even.push_back(a[i]);
        else odd.push_back(a[i]);
    }
    sort(even.begin(),even.end());
    sort(odd.begin(),odd.end());
    ll ecnt=1,ocnt=1;
    ll er=1,cr=1;
    ll ex=0,ox=0;
    for(ll i=1;i<even.size();i++){
        if(even[i-1]==even[i]){
            er++;
            if(er>ecnt) ex=i;
            ecnt=max(er,ecnt);
            continue;
        }
        er=1;
    }
    for(ll i=0;i<odd.size();i++){
        if(odd[i-1]==odd[i]){
            cr++;
            if(cr>ocnt) ox=i;
            ocnt=max(cr,ocnt);
            continue;
        }
        cr=1;
    }
    if(odd[ox]==even[ex]){
        cr=1,er=1;
        ll ocnt2=0,ecnt2=0;
        for(ll i=1;i<even.size();i++){
            if(even[i-1]==even[i]&&even[i]!=even[ex]){
                er++;
                ecnt2=max(er,ecnt2);
                continue;
            }
            if(even[i-1]==even[i]&&even[ex]==even[i]) continue;
            ecnt2=max(ecnt2,er);
            er=1;
        }
        for(ll i=1;i<odd.size();i++){
            if(odd[i-1]==odd[i]&&odd[i]!=odd[ox]){
                cr++;
                ocnt2=max(cr,ocnt2);
                continue;
            }
            if(odd[i-1]==odd[i]&&odd[i]==odd[ox]) continue;
            ocnt2=max(ocnt2,cr);
            cr=1;
        }
        if(ocnt+ecnt2>ocnt2+ecnt) ecnt=ecnt2;
        else ocnt=ocnt2;
    }
    ll ans=n-ecnt-ocnt;
    cout << ans << endl;
}