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 <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <utility>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
using ll = long long;
using itn = int;
using namespace std;
int GCD(int a, int b){
    return b ? GCD(b, a%b) : a;
}
int main(){
    int n;
    cin >> n;
    int odd[n/2],even[n/2];
    for(int i=0; i<n; i++){
        int a;
        cin >> a;
        if(i%2){
            odd[i/2]=a;
        }else{
            even[i/2]=a;
        }
    }
    //奇数番目に現れる最大の数(maxOdd)と偶数番目に現れる最大の数(maxEven)でもって書き換える
    //書き換える必要がある個数はn-maxOdd-maxEvenで求められる気がする（ほんとか？
    //ただし最大の数が同じなら奇数で2番目に現れる最大の数（secondOdd）かその逆で書き換える（小さいほう。
    int maxOdd=0, maxEven=0, secondOdd=0, secondEven=0;
    int oi=-1,ei=-1;
    int memoOdd[100001]={}, memoEven[100001]={};
    for(int i=0; i<n/2; i++){
        memoOdd[odd[i]]++;
        memoEven[even[i]]++;
    }
    for(int i=0; i<100001; i++){
        if(maxOdd < memoOdd[i]){
            secondOdd = maxOdd;
            maxOdd = memoOdd[i];
            oi = i;
        }
        if(maxEven < memoEven[i]){
            secondEven = maxEven;
            maxEven = memoEven[i];
            ei = i;
        }
    }
    int ans=0;
    if(oi == ei){
        //secondの値を使う
        ans = min( (n-secondOdd-maxEven), (n-maxOdd-secondEven) );
    }else{
        ans = n-maxOdd-maxEven;
    }
    cout << ans << endl;

}