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>
#define MAX 100000

int v[MAX], c[2][MAX], V[MAX];

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

int Giza(int n){
  int i, j, k, m1, m2;
  int max[2], max2[2], mn[2], mn2[2];
  for(i = 0;i < n;i++){
    scanf("%d", &v[i]);
    V[i] = v[i];
  }
  qsort(V, n, sizeof(int), Cmp);
  for(i = 0;i < 2;i++) for(j = V[0];j <= V[n-1];j++) c[i][j] = 0;
  for(i = 0;i < n;i++) c[i%2][v[i]]++;
  for(i = 0;i < 2;i++){
    max[i] = max2[i] = mn[i] = mn2[i] = 0;
    for(j = V[0];j <= V[n-1];j++){
      if(max2[i] < c[i][j] || (i == 1 && max2[i] == c[i][j] && mn[0] != j && mn2[0] != j)){
	max2[i] = c[i][j];
	mn2[i] = j;
	if(max[i] < max2[i]){
	  k = max[i];
	  max[i] = max2[i];
	  max2[i] = k;
	  k = mn[i];
	  mn[i] = mn2[i];
	  mn2[i] = k;
	}
      }
    }
  }
  if(mn[0] != mn[1]) return n - (max[0] + max[1]);
  else{
    if(max[0] < max[1]){
      m1 = max[1];
      m2 = max2[0];
    }else{
      m1 = max[0];
      m2 = max2[1];
    }
    return n - (m1 + m2);
  }
}

int main(void){
  int n;
  scanf("%d", &n);
  printf("%d\n", Giza(n));
  return 0;
}