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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void){
  int n,v[100005],w[100005],elem1[100005]={},elem2[100005]={},i,k=0,l=0,max1,max2;
  int maxvalue1,maxvalue2,num1=0,num2=0,next,nextvalue,next1,nextvalue1,answer,minanswer;
  char str[1000000],*p;
  
  fgets(str,sizeof(str),stdin);
  n=atoi(str);
  
  fgets(str,sizeof(str),stdin);
  p=strtok(str," \n");
  v[l]=atoi(p),l++,num1=v[0];
  for(i=1;i<n;i++){
    p=strtok(NULL," \n");
    if(i%2==1){
      w[k]=atoi(p);
      if(num2<w[k]){
        num2=w[k];
      }
      k++;
    }else{
      v[l]=atoi(p);
      if(num1<v[l]){
        num1=v[l];
      }
      l++;
    }
  }
  
  for(i=0;i<n/2;i++){
    elem1[v[i]]++;
  }
  for(i=0,max1=0,next1=0,nextvalue1=0;i<n/2;i++){
    if(max1<=elem1[v[i]]){
      next1=max1,nextvalue1=maxvalue1,max1=elem1[v[i]],maxvalue1=v[i];
    }
  }
  for(i=0;i<=n/2;i++){
    elem2[w[i]]++;
  }
  for(i=0,max2=0,next=0,nextvalue=0;i<=n/2;i++){
    if(max2<elem2[w[i]]){
      next=max2,nextvalue=maxvalue2,max2=elem2[w[i]],maxvalue2=w[i];
    }else if(max2<=elem2[v[i]] && v[i]!=maxvalue1){
      next=max2,nextvalue=maxvalue2,max2=elem2[w[i]],maxvalue2=w[i];
    }
    
  }
  
  if(maxvalue1!=maxvalue2){
    answer=n-max1-max2;
    minanswer=answer;
  }
  if(maxvalue1==maxvalue2){
    answer=n-max1-next;
    minanswer=answer;
  }
  if(maxvalue1==maxvalue2 && nextvalue1!=maxvalue2){
    answer=n-next1-max2;
    if(answer<minanswer){
      minanswer=answer;
    }
  }
  
  printf("%d\n",minanswer);
  
  return 0;
}