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;
 int v[100005];
 int odd[100005];
 int even[100005];
 int i;
 int maxvalodd=0;
 int maxvaleven=0;
 int maxeven=-1;
 int maxevenindex=-1;
 int maxevenold=-1;
 int maxevenindexold=-1;
 int maxodd=-1;
 int maxoddindex=-1;
 int maxoddold=-1;
 int maxoddindexold=-1;
 int change;
 int changeodd;
 int changeeven;
 
 for(i=0;i<100005;i++)
 {
  even[i]=0;
  odd[i]=0;
 }

 scanf("%d",&n);
 for(i=0;i<n;i++) 
 {
  scanf("%d",&v[i]);
  if((i%2)==0)
  { 
   even[v[i]]++;
   if(v[i]>maxvaleven) maxvaleven=v[i];
  }else{
   odd[v[i]]++;
    if(v[i]>maxvalodd) maxvalodd=v[i];
  }
 }

 for(i=0;i<=maxvaleven;i++)
 {
  if(even[i]>=maxeven)
  {
   maxevenold=maxeven;
   maxevenindexold=maxevenindex;
   maxeven=even[i];
   maxevenindex=i;
  }
 }

 for(i=0;i<=maxvalodd;i++)
 {
  if(odd[i]>=maxodd) 
  {
   maxoddold=maxodd;
   maxoddindexold=maxoddindex;
   maxodd=odd[i];
   maxoddindex=i;
  }
 }

 if(maxoddindex==maxevenindex)
 {
  if(maxodd>maxeven)
  {
   maxeven=maxevenold;
   if(maxeven==-1) maxeven=0;
  }else{
   maxodd=maxoddold;
   if(maxodd==-1) maxodd=0;
  }
 }

 changeodd=n/2 - maxodd;
 changeeven=n/2 - maxeven;
 change=changeodd+changeeven;

 printf("%d\n",change);
 return 0;
}