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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define max(p,q)(((p)<(q)?(p):(q)))
#define min(p,q)(((p)>(q)?(p):(q)))

int main(void){
  long N;
  scanf("%ld",&N);

  if(N == 2){
    printf("0\n");
  }else{
    long odd[100000];
    long even[100000];
    long v;

    for(int i = 0;i < 100000;i++){
      odd[i] = 0;
      even[i] = 0;
    }
    for(int i = 0;i < N;i++){
      scanf("%ld",&v);
      if(i%2 == 0){
        even[v-1]++;
      }else{
        odd[v-1]++;
      }
    }

    long maxoddnum[2] = {max(odd[0],odd[1]),min(odd[0],odd[2])};
    long maxevennum[2] = {max(even[0],even[1]),min(even[0],even[2])};
    long oddnum = 0;
    long evennum = 0;

    for(int i = 1;i < 100000;i++){
      if(maxoddnum[0] < odd[i]){
        maxoddnum[1] = maxoddnum[0];
        maxoddnum[0] = odd[i];
        oddnum = i;
      }
      if(maxevennum[0] < even[i]){
        maxevennum[1] = maxevennum[0];
        maxevennum[0] = even[i];
        evennum = i;
      }
    }

    if(oddnum != evennum){
      printf("%ld\n",N-maxoddnum[0]-maxevennum[0]);
    }else if(maxoddnum[1] < maxevennum[1]){
      printf("%ld\n",N-maxoddnum[0]-maxevennum[1]);
    }else{
      printf("%ld\n",N-maxoddnum[1]-maxevennum[0]);
    }
  }
  return 0;
}