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 main(){
  int N;
  scanf("%d", &N);

  int O[N/2 + 1], E[N/2 + 1];
  O[N/2] = 0;
  E[N/2] = 0;

  int i, j;

  for(i = 0; i < N; i++){
    if(i%2 == 1) scanf("%d", &O[i/2]);
    else scanf("%d", &E[i/2]);
  }

  qsort(O, N/2, sizeof(int), compare_int);
  qsort(E, N/2, sizeof(int), compare_int);
  
  int max[2][2][2] = {};

  int count_O = 1, count_E = 1;

  for(i = 0; i < N/2; i++){
    if(O[i] != O[i+1]){
      if(max[0][0][0] < count_O){
        max[0][1][0] = max[0][0][0];
        max[0][1][1] = max[0][0][1];
        max[0][0][0] = count_O;
        max[0][0][1] = O[i];
      } else if(max[0][1][0] < count_O){
        max[0][1][0] = count_O;
        max[0][1][1] = O[i];
      }
      count_O = 1;
    } else count_O++;

    if(E[i] != E[i+1]){
      if(max[1][0][0] < count_E){
        max[1][1][0] = max[1][0][0];
        max[1][1][1] = max[1][0][1];
        max[1][0][0] = count_E;
        max[1][0][1] = E[i];
      } else if(max[1][1][0] < count_E){
        max[1][1][0] = count_E;
        max[1][1][1] = E[i];
      }
      count_E = 1;
    } else count_E++;
  }
  int ans;
  if(max[0][0][1] == max[1][0][1]){
    if(max[0][0][0] < max[1][0][0]){
      ans = N - max[1][0][0] - max[0][1][0];
    } else ans = N - max[0][0][0] - max[1][1][0];
  } else ans = N - max[0][0][0] - max[1][0][0];

  printf("%d\n", ans);

  return 0;
}