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>

int main(){
  int n;
  scanf("%d\n",&n);
  //
  if(n==2){
    n = 0;
    printf("%d",n);
    return 0;
  }
  //
  int m = n/2;
  int array1[m][2];
  int array2[m][2];
  for(int i=0;i<m;i++){
    array1[i][0] = 0;
    array2[i][0] = 0;
  }
  int x;
  int last1 = 0;
  int last2 = 0;
  scanf("%d%*c",&x);
  array1[0][0] = x;
  array1[0][1] = 1;
  scanf("%d%*c",&x);
  array2[0][0] = x;
  array2[0][1] = 1;
  //
  int j,k;
  for(int i=2;i<n;i++){
    scanf("%d%*c",&x);
    j = 0;
    k = i%2;
    while(1){
      if(k==0){
	if(array1[j][0] == 0){
	  array1[j][0] = x;
	  array1[j][1] = 1;
	  last1++;
	  break;
	}else if(array1[j][0] == x){
	  array1[j][1] += 1;
	  break;
	}else{
	  j++;
	}
      }else{
	if(array2[j][0] == 0){
	  array2[j][0] = x;
	  array2[j][1] = 1;
	  last2++;
	  break;
	}else if(array2[j][0] == x){
	  array2[j][1] += 1;
	  break;
	}else{
	  j++;
	}
      }
    }
  }
  //
  if(array1[0][0]==array2[0][0] && last1==0 && last2==0){
    m = n/2;
    printf("%d",m);
    return 0;
  }
  //
  int answer,max,amount,count;
  answer = 0;
  max = array1[0][1];
  amount = array1[0][1];
  count = 1;
  while(last1){
    if(max < array1[count][1]){
      max = array1[count][1];
      answer += amount;
      amount = array1[count][1];
      count++;
      last1--;
    }else{
      answer += array1[count][1];
      count++;
      last1--;
    }
  }
  max = array2[0][1];
  amount = array2[0][1];
  count = 1;
  while(last2){
    if(max < array2[count][1]){
      max = array2[count][1];
      answer += amount;
      amount = array2[count][1];
      count++;
      last2--;
    }else{
      answer += array2[count][1];
      count++;
      last2--;
    }
  }
  printf("%d",answer);
  return 0;
}