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: Accepted)

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

typedef struct {
  int num;
  int mass;
} bigint;

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

int compare_bigint( const void *p, const void *q ) {
   return ((bigint*)q)->mass - ((bigint*)p)->mass;
}

int min(int a,int b){
  if(a<b){
    return a;
  }
  return b;
}

void calic(int *A,int n,bigint*ma,int*a){
  int i;
  int k=1;
  int j=0;
  for(i=0;i<n;i++){
    ma[i].mass=0;
    ma[i].num=0;
  }
  for(i=1;i<n/2;i++){
    if(A[i]==A[i-1]){
      k++;
    }else{
      ma[j].mass=k;
      ma[j].num=A[i-1];
      j++;
      k=1;
    }
  }
  ma[j].mass=k;
  ma[j].num=A[n/2-1];
  a[0]=j+1;
}

int main(){
  int n,i,ans;
  scanf("%d",&n);
  int A[n/2];
  int B[n/2];
  for(i=0;i<n;i++){
    if(i%2==0){
      scanf("%d",&A[i/2]);
    }else{
      scanf("%d",&B[(i-1)/2]);
    }
  }

  bigint ma[n];
  bigint mb[n];
  int a[1];
  int b[1];
  qsort(A, n/2, sizeof(int), compare_int);
  qsort(B, n/2, sizeof(int), compare_int);
  calic(A,n,ma,a);
  calic(B,n,mb,b);
  qsort(ma, a[0], sizeof(bigint), compare_bigint);
  qsort(mb, b[0], sizeof(bigint), compare_bigint);

  
  if(ma[0].num==mb[0].num){
    ans=min(n-ma[1].mass-mb[0].mass,n-ma[0].mass-mb[1].mass);
  }else{
    ans=n-(ma[0].mass+mb[0].mass);
  }
  printf("%d",ans);
  return 0;
}