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

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

void calic(int *A,int n,int *ma){
  int i;
  int k=1;
  int flag=0;
  for(i=1;i<n/2;i++){
    if(A[i]==A[i-1]){k++;}
    if(A[i]!=A[i-1] && ma[0]<=k){
      if(flag!=0){ma[1]=ma[0];}
      ma[0]=k;flag=1;ma[2]=A[i-1];k=1;
    }
    if(A[i]!=A[i-1] && ma[0]>k && k>ma[1]){
      ma[1]=k;k=1;flag=1;
    }
  }
  
  if(ma[0]<=k){
    if(flag!=0){ma[1]=ma[0];}
    ma[0]=k;ma[2]=A[n/2-1];
  }
  if(ma[0]>k&&k>ma[1]){
      ma[1]=k;
  }
}

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]);
    }
  }
  qsort(A, n/2, sizeof(int), compare_int);
  qsort(B, n/2, sizeof(int), compare_int);
  int ma[3]={1,0,A[0]};
  int mb[3]={1,0,B[0]};
  calic(A,n,ma);
  calic(B,n,mb);
  
  if(mb[2]==ma[2]){
    ans=min(n-ma[0]-mb[1],n-ma[1]-mb[0]);
  }else{
    ans=n-(ma[0]+mb[0]);
  }
  printf("%d",ans);
  return 0;
}