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>

main()
{
  int n,*a,*b,i,maxa=0,maxai=0,maxb=0,maxbi=0,count;
  int maxa2=0,maxai2=0,maxb2=0,maxbi2=0;
  int c[1000000]={};
  scanf("%d",&n);
  a = malloc(sizeof(int)*(n/2));
  b = malloc(sizeof(int)*(n/2));
  for(i=0; i<n/2; i++)
  {
    scanf("%d%d",&a[i],&b[i]);
  }
//aの処理  
  for(i=0; i<n/2; i++)
  {
    c[a[i]]++;
    if(maxa<=c[a[i]])
    {
      maxa = c[a[i]];
      maxai = a[i];
    }
  }
  for(i=0; i<n/2; i++)
  {
    if(maxa2<=c[a[i]]&&maxa!=c[a[i]])
    {
      maxa2 = c[a[i]];
      maxai2 =  a[i];
    }
  }
//bの処理
  for(i=0; i<1000000; i++)
    c[i] = 0;
  for(i=0; i<n/2; i++)
  {
    c[b[i]]++;
    if(maxb<=c[b[i]])
    {
      if(maxbi!=b[i])
      maxb = c[b[i]];
      maxbi = b[i];
    }
  }
  for(i=0; i<n/2; i++)
  {
    if(maxb2<=c[b[i]]&&maxb!=c[b[i]])
    {
      maxb2 = c[b[i]];
      maxbi2 =  b[i];
    }
  }
  count = n - maxa - maxb;
  if(count!=0)
  {
      if(maxai!=maxbi)
        printf("%d\n",count);
      else if(maxa>=maxb)
        printf("%d\n",n-maxa-maxb2);
      else if(maxa<maxb)
        printf("%d\n",n-maxa2-maxb);
  }
  else
  {
    if(a[0]==b[0])
      printf("%d\n",n/2);
    else
      printf("0\n");
  }
}