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>
#include <stdlib.h>

int compare_int(const void *a,const void *b){
  if(*(long*)a-*(long*)b > 0) return 1;
  return -1;
}

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

  if(N == 2){
    int a,b;
    scanf("%d%d",&a,&b);
    if(a != 0) printf("0\n");
    else printf("1\n");
    return 0;
  }

  N /= 2;

  int odd[N];
  int even[N];
  for(int i = 0;i < N;i++){
    scanf("%d%d",&odd[i],&even[i]);
  }

  qsort(odd,N,sizeof(int),compare_int);
  qsort(even,N,sizeof(int),compare_int);

  int oddlen1 = 0,oddlen2 = 0;
  int evenlen1 = 0,evenlen2 = 0;

  int cnt = 0;
  int past = odd[0];
  int oddnum = odd[0];
  for(int i = 0;i < N;i++){
    if(odd[i] == past && i != N-1){
      cnt++;
    }else{
      if(i == N-1 && odd[i] == past) cnt++;
      if(oddlen1 <= cnt){
        oddnum = past;
        oddlen2 = oddlen1;
        oddlen1 = cnt;
      }
      past = odd[i];
      cnt = 0;
    }
  }
  cnt = 0;
  past = even[0];
  int evennum = even[0];
  for(int i = 0;i < N;i++){
    if(even[i] == past && i != N-1){
      cnt++;
    }else{
      if(i == N-1 && even[i] == past) cnt++;
      if(evenlen1 <= cnt){
        evennum = past;
        evenlen2 = evenlen1;
        evenlen1 = cnt;
      }
      past = even[i];
      cnt = 0;
    }
  }

  if(oddnum != evennum){
    printf("%d\n",2*N-oddlen1-evenlen1);
  }else if(oddlen1+evenlen2 > oddlen2+evenlen1){
    printf("%d\n",2*N-oddlen1-evenlen2);
  }else{
    printf("%d\n",2*N-oddlen2-evenlen1);
  }

  return 0;
}