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>
int compare(const void *a,const void *b){
  return *(int*)a-*(int*)b;
}

int main(){
  int a[100000] = {};
  int b[100000] = {};
  int n=0;
  int i=0;
  int maxa=0;
  int maxb=0;
  int seca=0;
  int secb=0;
  int mackya = 0;
  int mackyb = 0;
  int seckya = 0;
  int seckyb = 0;
  int cnt = 0;
  int hozon = 0;
  
  scanf("%d",&n);
  
  for(i=0;i<n/2;i++){
    scanf("%d %d",&a[i],&b[i]);
  }
  
  qsort(a,n/2,sizeof(int),compare);
  qsort(b,n/2,sizeof(int),compare);
  hozon=a[0];
  cnt = 1;
  
  for(i=1;i<n/2;i++){
    if(a[i] == hozon){
      cnt++;
    }else{
      if(cnt >= maxa){
        seca = maxa;
        maxa = cnt;
        seckya = mackya;
        mackya = hozon;
      }else if(cnt > seca){
        seca = cnt;
        seckya = hozon;
      }
      cnt = 1;
      hozon = a[i];
    }
  }
  
  if(cnt >= maxa){
    seca = maxa;
    maxa = cnt;
    seckya = mackya;
    mackya = hozon;
  }else if(cnt > seca){
    seca = cnt;
    seckya = hozon;
  }    
  //
  
  hozon=b[0];
  cnt = 1;
  
  for(i=1;i<n/2;i++){
    if(b[i] == hozon){
      cnt++;
    }else{
      if(cnt >= maxb){
        seca = maxb;
        maxb = cnt;
        seckyb = mackyb;
        mackyb = hozon;
      }else if(cnt > secb){
        secb = cnt;
        seckyb = hozon;
      }
      cnt = 1;
      hozon = b[i];
    }
  }
  
  if(cnt >= maxb){
    secb = maxb;
    maxb = cnt;
    seckyb = mackyb;
    mackyb = hozon;
  }else if(cnt > secb){
    secb = cnt;
    seckyb = hozon;
  } 
  
  if(seckya == 0){
    seckya = mackya;
  }
  if(seckyb == 0){
    seckyb = mackyb;
  }
  //printf("%d %d %d %d %d %d\n",mackya,seckya,mackyb,seckyb,maxa,maxb);
  if(mackya != mackyb){
    printf("%d",n-maxa-maxb);
  }else{
    if(maxa+secb > maxb+seca){
      printf("%d",n-maxa-secb);
    }else{
      printf("%d",n-maxb-seca);
    }
  }
  
  return 0;
}