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 <string.h>
#define MAX 100000

int main() {
  long int n;

  long int i, tmp, x, max_odd, max_odd_i, sec_odd, max_even, max_even_i,
      sec_even, y, z;
  long int odd[MAX] = {0}, even[MAX] = {0};

  scanf("%ld", &n);

  for (i = 0; i < n; i++) {
    scanf("%ld", &tmp);
    if (i % 2 == 0) {
      even[tmp - 1]++;
    } else {
      odd[tmp - 1]++;
    }
  }

  max_odd = 0;
  max_odd_i = 0;
  sec_odd = 0;
  for (i = 0; i < MAX; i++) {
    if (odd[i] >= max_odd) {
      sec_odd = max_odd;
      max_odd = odd[i];
      max_odd_i = i;
    }
  }
  max_even = 0;
  max_even_i = 0;
  sec_even = 0;
  for (i = 0; i < MAX; i++) {
    if (even[i] >= max_even) {
      sec_even = max_even;
      max_even = even[i];
      max_even_i = i;
    }
  }

  if (max_odd_i != max_even_i) {
    x = n - max_odd - max_even;
  } else {
    // max_oddにする
    y = n - max_odd - sec_even;
    // max_evenにする
    z = n - sec_odd - max_even;
    x = y < z ? y : z;
  }
  printf("%ld\n", x);
  return 0;
}