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>
#include <unistd.h>
#include <string.h>
#include <stdio.h>

int count(int a[],int k,int n);
void sort(int numbers[], int left, int right);

int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  int m = n/2;
  int a[m];
  int b[m];
  int amax = 0;
  int bmax = 0;
  int acount = 0;
  int bcount = 0;
  int ans = 0;
  for (int i = 0; i < m; i++) {
    scanf("%d", &a[i]);
    scanf("%d", &b[i]);
  }
  sort(a,0,m-1);
  sort(b,0,m-1);

  int temp = 0;
  for (int i = 0; i < m; i++) {
    if(i == 0){
      amax = a[i];
      temp = 1;
    }else if(a[i-1] == a[i]){
      temp++;
    }else{
      if(temp > acount){
        acount = temp;
        amax = a[i-1];
      }
      temp = 1;
    }
  }
  if(temp > acount){
    acount = temp;
    amax = a[m-1];
  }

  temp = 0;
  for (int i = 0; i < m; i++) {
    if(i == 0){
      bmax = b[i];
      temp = 1;
    }else if(b[i-1] == b[i]){
      temp++;
    }else{
      if(temp > bcount){
        bcount = temp;
        bmax = b[i-1];
      }
      temp = 1;
    }
  }
  if(temp > bcount){
    bcount = temp;
    bmax = b[m-1];
  }

  // temp = 0;
  // for (int i = 0; i < m; i++) {
  //   temp = count(b,b[i],m);
  //   if(temp > bcount){
  //     bmax = b[i];
  //     bcount = temp;
  //   }
  // }
  if(amax == bmax){
    ans = m;
  }else{
    ans = n - acount - bcount;
  }
  printf("%d",ans);
  return 0;
}

int count(int a[],int k,int n){
  int count = 0;
  for (int i = 0; i < n; i++) {
    if(a[i] == k){
      count++;
    }
  }
  return count;
}

void sort(int numbers[], int left, int right)
{
    int pivot, l_hold, r_hold;

    l_hold = left;
    r_hold = right;
    pivot = numbers[left];
    while (left < right)
    {
        while ((numbers[right] >= pivot) && (left < right))
            right--;
        if (left != right)
        {
            numbers[left] = numbers[right];
            left++;
        }
        while ((numbers[left] <= pivot) && (left < right))
            left++;
        if (left != right)
        {
            numbers[right] = numbers[left];
            right--;
        }
    }
    numbers[left] = pivot;
    pivot = left;
    left = l_hold;
    right = r_hold;
    if (left < pivot)
        sort(numbers, left, pivot-1);
    if (right > pivot)
        sort(numbers, pivot+1, right);
}