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 <math.h>
#include <stdlib.h>

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


int main(){
    int a,b;
    int max_ev=0,max_od=0,sec_ev=0,sec_od=0;
    int sum1,sum2,sum3;
    int even[100001]={};
    int odd[100001]={};
    
    scanf("%d",&a);
    for(int i=1;i<a+1;i++){
        if(i%2==1){
            scanf("%d",&b);
            odd[b]+=1;
        }
        else{
            scanf("%d",&b);
            even[b]+=1;
        }
    }
    //puts("__________");
    
    qsort(even,100001,sizeof(even[0]),int_sort);
    qsort(odd,100001,sizeof(odd[0]),int_sort);
    max_ev=even[0];max_od=odd[0];
    sec_ev=even[1];sec_od=odd[1];
    
    sum1=a-(max_ev+max_od);
    sum2=a-(max_ev+sec_od);
    sum3=a-(sec_ev+max_od);
   // printf("------\n%d %d %d\n",sum1,sum2,sum3);
    
    if(sum1>=sum3 &&sum2>=sum3){
        printf("%d\n",sum3);
        return 0;
    }
    
    if(sum1>=sum2 &&sum3>=sum2){
        printf("%d\n",sum2);
        return 0;
    }
    if(sum3>=sum1 &&sum2>=sum1){
        printf("%d\n",sum1);
        return 0;
    }
    return 0;
}