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>
#include <math.h>
long long int gcd(long long int a,long long int b);

int main (void){
int a[100001],b[100001],n,tmp;
int g_first=0,g_second=0,g_first_n=0,g_second_n=0;
int k_first=0,k_second=0,k_first_n=0,k_second_n=0;
for (int i = 0; i <= 100000; i++) {
  a[i]=0;
  b[i]=0;
}
scanf("%d\n",&n);
if (n==2) {
  int A,B;
  scanf("%d\n",&A );
  scanf("%d\n",&B );
  if (A==B) {
    printf("1\n" );
  }else{
    printf("0\n" );
  }
  return 0;
}
for (int i = 1; i <= n; i++) {
  scanf("%d\n", &tmp);
  if (i%2==1) {
      a[tmp]++;
  }else{
    b[tmp]++;
  }

}

for (int i = 1; i <= 100000; i++) {
  if (a[i]>=k_first_n) {
    k_second_n=k_first_n;
    k_second=k_first;
    k_first_n=a[i];
    k_first=i;
  }else{
    if (a[i]>=k_second_n) {
      k_second_n=a[i];
      k_second=i;
    }
  }
}
for (int i = 1; i <= 100000; i++) {
  if (b[i]>=g_first_n) {
    g_second_n=g_first_n;
    g_second=g_first;
    g_first_n=b[i];
    g_first=i;
  }else{
    if (b[i]>=g_second_n) {
      g_second_n=b[i];
      g_second=i;
    }
  }
}
//printf("%d\n",g_first );
//printf("%d\n",k_first );
if (g_first!=k_first) {
  printf("%d\n", n-g_first_n-k_first_n);
}else{
  if (g_second_n>=k_second_n) {
    printf("%d\n", n-g_second_n-k_first_n);
  }else{
    printf("%d\n", n-g_first_n-k_second_n);
  }
}
  return 0;
}