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: Wrong Answer)

#include <iostream>
#include <string>

using namespace std;

int const NMAX = 1e5;
int freqEv[1 + NMAX + 1];
int freqOd[1 + NMAX + 1];

int main() {

  int n, m, ans = 0, isB = 0, Od = NMAX + 1, Ev = NMAX + 1;
  string p;
  cin >> n;
  for(int i =1;i <= n;i++){
    cin >> isB;
    if(i % 2 == 1){
      freqOd[isB]++;
      if(freqOd[freqOd[0]] < freqOd[isB]){
        if(freqOd[0] != 0){
          Od = freqOd[0];
        }
        freqOd[0] = isB;
      }else if(freqOd[Od] < freqOd[isB] && freqOd[0] != isB){
        Od = isB;
      }
    }else{
      freqEv[isB]++;
      if(freqEv[freqEv[0]] < freqEv[isB]){
        if(freqEv[0] != 0){
          Ev = freqEv[0];
        }
        freqEv[0] = isB;
      }else if(freqEv[Ev] < freqEv[isB] && freqEv[0] != isB){
        Ev = isB;
      }
    }
  }
  //cerr << Ev << ' ' << Od;
  if(freqEv[0] == freqOd[0]){
    cout << max(n - freqEv[Ev] - freqOd[freqOd[0]], n - freqEv[freqEv[0]] - freqOd[Od]);
  }else{
    cout << n - freqEv[freqEv[0]] - freqOd[freqOd[0]];
  }
  return 0;
}