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 9.2.1) (judge verdict: Wrong Answer)

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

int down(const void *a,const void *b){
  return *(int *)b - *(int *)a;
}

int main(void){
  int n;
  scanf("%d",&n);
  
  int v[n+1] , kisu[100001] = {0} , gusu[100001] = {0};
  int kisumax = 0 , gusumax = 0;
  int x , y;

  for(int i = 1;i <= n;i ++){
    scanf("%d",&v[i]);
    if(i%2 == 1){
      kisu[v[i]] ++;
      if(kisumax < kisu[v[i]]){
        kisumax = kisu[v[i]];
        x = v[i];
      }
    }

    else{
     gusu[v[i]] ++;
     if(gusumax < gusu[v[i]]){
       gusumax = gusu[v[i]];
       y = v[i];
     }
    }
  }

  int ans = 0;
  if(x != y){
    for(int i = 1;i < 100001;i ++){
    ans += kisu[i] + gusu[i];
  }
  ans -= kisumax + gusumax;
  }

  else{
    qsort(kisu,100001,sizeof(int),down);
    qsort(gusu,100001,sizeof(int),down);

    for(int i = 0;i < 100001;i ++){
      ans += kisu[i] + gusu[i];
    }
    if(kisumax > gusumax){
      ans -= kisumax + gusu[1];
    }
    else{
      ans -= gusumax + kisu[1];
    }
  }
  
  printf("%d",ans);

  return 0;
}