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: Runtime Error)

#include <stdio.h>
int main(void){
  int num;
  scanf("%d\n", &num);
  int N[num];
  int i=0;
  int j=0;
  int k=0;
  int distance;
  int Nrank[3]={0};//this has exact number of [top 3]
  int Ncount[num];//this has count of each [exact number]
  for(i=0; i<num;i++){
    Ncount[i]=0;
    N[i]=0;
  }
  while(scanf("%d", &N[i])!=EOF){
    printf("%d\n", N[i]);
    Ncount[N[i]]++;
    for (j=0; j<3;j++){
      if ((Ncount[Nrank[j]]<Ncount[N[i]])&&(Nrank[j]!=N[i])){
        if(j==2){
          Nrank[j]=N[i];
          break;
        }
        if(j==1){
          Nrank[j+1]=Nrank[j];
          Nrank[j]=N[i];
          break;
        }
        else{
          Nrank[j+2]=Nrank[j+1];
          Nrank[j+1]=Nrank[j];
          Nrank[j]=N[i];
          break;
        }
   	 }//end of if statement
    }//end of for statement

    i++;
 	}

    distance = num - (Ncount[Nrank[0]]+Ncount[Nrank[1]]);
    if (distance==0){
      if (Ncount[Nrank[1]]==0){
        printf("%d", num/2);
      }
      else{
        printf("%d", Ncount[Nrank[0]]-Ncount[Nrank[1]]);
      }
    }
    else{
      printf("%d", distance);
    }
  return 0;
}