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>
int main(void){
    int n,max=0;
    scanf("%d",&n);
    int max_od=0,top2_od=0,max_ev=0,top2_ev=0;
    int v[n];
    int v_od[100001]={0},v_ev[100001]={0};
    //
    for(int i=0;i<n;i++){
        scanf("%d",&v[i]);
        if(max<v[i])max=v[i];
        if(i%2==0){
            v_od[v[i]]++;
        }else{
            v_ev[v[i]]++;
        }
    }

    /*    
    for(int i=0;i<=max;i++){
        printf("v_od[%d]:%d  v_ev[%d]:%d\n",i,v_od[i],i,v_ev[i]);
    }
    */
    

    for(int i=1;i<=max;i++){
        if(v_od[i]>v_od[max_od]){
            max_od=i;
        }else if(v_od[i]>v_od[top2_od]){
            top2_od=i;
        }
    }
    for(int i=1;i<=max;i++){
        if(v_ev[i]>=v_ev[max_ev]){
            max_ev=i;
        }else if(v_ev[i]>v_ev[top2_ev]){
            top2_ev=i;
        }
    }
    if(max_od!=max_ev){
        printf("%d\n",n-v_od[max_od]-v_ev[max_ev]);
    }else{
        //if((n-v_od[max_od]-v_ev[top2_ev]) > (n-v_od[top2_od]-v_ev[max_ev])){
        if(v_od[top2_od] > v_ev[top2_ev]){
            printf("%d\n",n-v_od[top2_od]-v_ev[max_ev]);
        }else{
            printf("%d\n",n-v_od[max_od]-v_ev[top2_ev]);
        }
    }
    /*
    printf("od_max:%d  od_top2:%d\nev_max:%d  ev_top2:%d\n",v_od[max_od],v_od[top2_od],v_ev[max_ev],v_ev[top2_ev]);
    printf("max:%d\n",max);
    printf("%d\n",v_ev[87735]);
    */
    return(0);

}