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 limit 100010

int odd[limit];
int even[limit];

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

  N /= 2;

  for(int i = 0;i < limit;i++){
    odd[i] = even[i] = 0;
  }
  int a,b;
  for(int i = 0;i < N;i++){
    scanf("%d%d",&a,&b);
    odd[a-1]++;
    even[b-1]++;
  }

  if(N == 1){
    if(a != 0) printf("0\n");
    else printf("1\n");
    return 0;
  }

  int oddlen1 = 0,oddlen2 = 0;
  int evenlen1 = 0,evenlen2 = 0;
  int oddnum = a,evennum = b;

  for(int i = 0;i < limit;i++){
    if(oddlen1 <= odd[i]){
      oddlen2 = oddlen1;
      oddlen1 = odd[i];
      oddnum = i;
    }
    if(evenlen1 <= even[i]){
      evenlen2 = evenlen1;
      evenlen1 = even[i];
      evennum = i;
    }
  }

  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;
}