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: Runtime Error)

#include <math.h>
#include <stdio.h>

int main()
{
    int n, i, k, j, a, K, J;

    a = 0;

    scanf("%d", &n);

    K = n/2;
    J = (n+1)/2;

    int x[n];
    int A[K][J];
    
    for(i=0; i<n; i++){
      scanf("%d", &x[i]);
    }

    for(k=0; 2*k<n; k++){
      for(j=0; 2*j+1<n; j++){
        a = 0;
        if(x[2*k] == x[2*j+1]){
          x[2*j+1] += 1;
        }
          for(i=0; 2*i<n; i++){
            if(x[2*i] != x[2*k]){
              a += 1;
            }
          }
          for(i=0; 2*i+1<n; i++){
            if(x[2*i+1] != x[2*j+1]){
              a += 1;
            }
          }
        
        A[k][j] = a;
      }
    }

    int b;
    int B[K];

    for(k=0; k<K; k++){
      B[k] = A[k][0];
    }

    b = B[0];

    for(k=0; k<K; k++){
      for(j=0; j<J; j++){
        if(B[k]<A[k][j])
          B[k] = A[k][j];
      }
      if(b<B[k]){
        b = B[k];
      }
    }
    printf("%d\n", b);

    return 0;
}