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: Accepted)

#include<bits/stdc++.h>

using namespace std;

int main(){
    int N;
    int odd[100001];
    int even[100001];
    int v[100001];
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> v[i];
    }

    for(int i = 1;i < N; i+=2){
        odd[v[i]]++;
        even[v[i-1]]++;
    }

    int omax = 0,emax = 0, omax2 = 0, emax2 = 0;
    for(int i = 0;i < 100001; i++){
        if(omax < odd[i]){
            omax = odd[i];
            omax2 = i;
        }
        if(emax < even[i]){
            emax = even[i];
            emax2 = i;
        }
    }

    if(omax2 != emax2){
        cout << (N -omax -emax) << endl;
    }else{
        int osec = 0, esec = 0;
        for(int i = 0;i < 100001; i++){
            //なるほど、ほんとの最大値は引っかからないようにif文を設定して最大値をもう一度求めることで二番目の最大値を求めれるというわけか
            if(osec <= odd[i] && i!= omax2){
                osec = odd[i];
            }
            if(esec < even[i] && i != emax2){
                esec = even[i];
            }
        }
        cout << min((N-omax-esec),(N-emax-osec));
    }
    return 0;
}