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>

int main(void){
  int n,ans=0;
  int count[100000],num[100000],ptr=0;
  int odd[100000],even[100000];
  scanf("%d",&n);
  int i;
  for(i=0;i<n/2;i++){
    scanf("%d",&odd[i]);
    scanf("%d",&even[i]);
  }
  count[0]=1;
  num[0]=odd[0];
  ptr=1;
  int j;
  for(i=1;i<n/2;i++){
    int f=1;
    for(j=0;j<ptr;j++){
      if(num[j]==odd[i]){
        count[j]++;
        f=0;
      }
    }
    if(f){
      num[ptr]=odd[i];
      count[ptr]=1;
      ptr++;
    }
  }
  int max1=-1;
  for(i=0;i<ptr;i++){
    if(count[i]>max1) max1=count[i];
  }

  count[0]=1;
  num[0]=even[0];
  ptr=1;
    for(i=1;i<n/2;i++){
    int f=1;
    for(j=0;j<ptr;j++){
      if(num[j]==even[i]){
        count[j]++;
        f=0;
      }
    }
    if(f){
      num[ptr]=even[i];
      count[ptr]=1;
      ptr++;
    }
  }
  int max2=-1;
  for(i=0;i<ptr;i++){
    if(count[i]>max2) max2=count[i];
  }
  ans=n-max1-max2;
  if(ans==0 && odd[0]==even[0]){
    ans=n/2;
  }
  printf("%d\n",ans);

  return 0;
}