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: Accepted)

#include <stdio.h>

int main(){
  int n;
  int e,o;
  int even[100001];
  int odd[100001];
  int even_max=0,even_max_n=0;
  int even_second=0,even_second_n=0;
  int odd_max=0,odd_max_n=0;
  int odd_second=0,odd_second_n=0;
  
  for(int i=1;i<=100000;i++)
    even[i]=odd[i]=0;
  
  scanf("%d",&n);
  for(int i=0;i<n/2;i++){
    scanf("%d %d",&o,&e);
    odd[o]++;
    even[e]++;
  }
  
  for(int i=1;i<=100000;i++){
    if(even[i]>even_max){
      even_second=even_max;
      even_second_n=even_max_n;
      even_max=even[i];
      even_max_n=i;
    }else if(even[i]>even_second){
      even_second=even[i];
      even_second_n=i;
    }
    if(odd[i]>odd_max){
      odd_second=odd_max;
      odd_second_n=odd_max_n;
      odd_max=odd[i];
      odd_max_n=i;
    }else if(odd[i]>odd_second){
      odd_second=odd[i];
      odd_second_n=i;
    }
  }

  if(odd_max_n!=even_max_n)
    printf("%d\n",n-odd_max-even_max);
  else
    if(even_second<odd_second)
      printf("%d\n",n-odd_second-even_max);
    else
      printf("%d\n",n-odd_max-even_second);

  return 0;
}