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 <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <numeric>
#include <tuple>

// さんこう :https://beta.atcoder.jp/contests/abc109/submissions/3153889
using namespace std;

typedef long long ll;

int N;
int v[100010];
int dp[2][100010];

int main() {
   cin >> N;
   for (int i=0;i<N;i++) {
      cin >> v[i];
      dp[i % 2][v[i]]++;
   }
   int max[2] = {100010, 100020};
   int max_2nd[2] = {100009, 100008};
   int max_tmp[2] = {0, 0};
   for (int i=1;i<=100000;i++) {
      if (max_tmp[0] < dp[0][i]) {
         max_tmp[0] = dp[0][i];
         max_2nd[0] = max[0];
         max[0] = i;
      }
      if (max_tmp[1] < dp[1][i]) {
         max_tmp[1] = dp[1][i];
         max_2nd[1] = max[1];
         max[1] = i;
      }
   }
   int should_change_basis[2] = {max[0], max[1]};
   if (should_change_basis[0] == should_change_basis[1]) {
      should_change_basis[1] = max_2nd[1];
      // もはや1種類しかない場合でもこれで問題ない
   }
   printf("%d\n", N - dp[0][should_change_basis[0]] - dp[1][should_change_basis[1]]);
   return 0;
}