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

int main(){
  int i,j;
  int l;
  int n;
  int p[100000];
  int max1=0;
  int max2=0;
  int max11,max22=0;
  int maxnum1;
  int maxnum2;
  int maxnum11;
  int maxnum22;
  int a,b=0;

  for(i=0; i<100000; i++){
    p[i]=0;
  }

  scanf("%d",&n);

  int v[n];

  for(i=0; i<=n-1; i++){
    scanf("%d ",&v[i]);
  }

  for(i=0; i<=n-1; i++){
    if(i%2==0){
      l = v[i];
      p[l]++;
    }
  }
  for(i=0; i<100000; i++){
    if(p[i]>max1){
      max1 = p[i];
      maxnum1 = i;
    }
  }
  for(i=0; i<100000; i++){
    if(p[i]>max11 && i!=maxnum1){
      max11 = p[i];
      maxnum11 = i;
    }
  }
  //printf("%d\n",maxnum1);
  for(i=0; i<100000; i++){
    p[i]=0;
  }
  for(i=0; i<=n-1; i++){
    if(i%2==1){
      l = v[i];
      p[l]++;
    }
  }
  for(i=0; i<100000; i++){
    if(p[i]>max2){
      max2 = p[i];
      maxnum2 = i;
    }
  }
  for(i=0; i<100000; i++){
    if(p[i]>max22 && i!=maxnum2){
      max22 = p[i];
      maxnum22 = i;
    }
  }

  if(maxnum1!=maxnum2){
    for(i=0; i<=n-1; i++){
      if(i%2==0){
        if(v[i]!=maxnum1){
          a++;
        }
      }else if(i%2==1){
        if(v[i]!=maxnum2){
          b++;
        }
      }
    }
  }
  if(maxnum1==maxnum2){
    if(max1+max22>=max2+max11){
      for(i=0; i<=n-1; i++){
        if(i%2==0){
          if(v[i]!=maxnum1){
            a++;
          }
        }else if(i%2==1){
          if(v[i]!=maxnum22){
            b++;
          }
        }
      }
    }else if(max1+max22>=max2+max11){
      for(i=0; i<=n-1; i++){
        if(i%2==0){
          if(v[i]!=maxnum11){
            a++;
          }
        }else if(i%2==1){
          if(v[i]!=maxnum2){
            b++;
          }
        }
      }
    }
  }

  //printf("%d\n",a);
  printf("%d\n",a+b);


  return 0;
}