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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

int int_sort( const void * a , const void * b ) {
  if( *( int * )a < *( int * )b ) {
    return -1;
  }
  else
    if( *( int * )a == *( int * )b ) {
      return 0;
    }
  return 1;
}


int main(void){

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

  int *A,*B;
  A = (int*)malloc((N/2)*sizeof(int));
  B = (int*)malloc((N/2)*sizeof(int));
  for(int i=0; i<N; i++){
    scanf("%d %d ", &A[i], &B[i]);
  }

  qsort(A,N/2,sizeof(int),int_sort);
  qsort(B,N/2,sizeof(int),int_sort);
  int num=1,maxA1=0,maxA2=0,keyA1=0,keyA2=0,maxB1=0,maxB2=0,keyB1=0,keyB2=0;

  for(int i=0; i<N/2-1; i++){
    if(A[i]==A[i+1]){
      num++;
    }else{
      if(num>maxA1){
        maxA2 = maxA1;
        maxA1 = num;
        keyA2 = keyA1;
        keyA1 = A[i];
      }else if(num>maxA2){
        maxA2 = num;
        keyA2 = A[i];
      }
      num = 1;
    }
  }
  if(num>maxA1){
    maxA2 = maxA1;
    maxA1 = num;
    keyA2 = keyA1;
    keyA1 = A[N/2-1];
  }else if(num>maxA2){
    maxA2 = num;
    keyA2 = A[N/2-1];
  }
  num=1;

  for(int i=0; i<N/2-1; i++){
    if(B[i]==B[i+1]){
      num++;
    }else{
      if(num>maxB1){
        maxB2 = maxB1;
        maxB1 = num;
        keyB2 = keyB1;
        keyB1 = B[i];
      }else if(num>maxB2){
        maxB2 = num;
        keyB2 = B[i];
      }
      num = 1;
    }
  }
  if(num>maxB1){
    maxB2 = maxB1;
    maxB1 = num;
    keyB2 = keyB1;
    keyB1 = B[N/2-1];
  }else if(num>maxB2){
    maxB2 = num;
    keyB2 = B[N/2-1];
  }

  int ans=0;

  if(keyA1!=keyB1){
    ans = N-maxA1-maxB1;
  }else if(maxA2>maxB2){
    ans = N-maxA2-maxB1;
  }else{
    ans = N-maxA1-maxB2;
  }

  printf("A:%d-%d,%d-%d B:%d-%d,%d-%d\n", keyA1,maxA1,keyA2,maxA2,keyB1,maxB1,keyB2,maxB2);
  printf("%d\n", ans);

  return 0;
}