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: Time Limit Exceeded)

#include<stdio.h>

int main(){
    int n,a[100000],b[100000],k,l,i,j,z,count[2];
    scanf("%d",&n);
    for( i=0;i<n;i++){
        if(i%2==0){
            scanf("%d",&a[i/2]);
        }else{
            scanf("%d",&b[i/2]);
    };
    }

    count[0]=0;
    count[1]=0;
for(i=0;i<n/2;i++){
    k=1;
    for(j=0;j<n/2;j++){
        if(a[i]==a[j]){
            k++;
        }
                    if(count[0]<k-1){
                count[0]=k-1;
                l=a[i];
            }
    }
    }

for(i=0;i<n/2;i++){
    k=1;
    for(j=0;j<n/2;j++){
        if(b[i]==b[j]){
            k++;
        }
                    if(count[1]<k-1){
                count[1]=k-1;
                z=b[i];
            }
    }
    }
    if(z!=l){
    printf("%d",n-count[1]-count[0]);
    }else{
        if(count[0]<count[1]){
                printf("%d",n-count[1]);
        }else{
                printf("%d",n-count[0]);
        }
    }
    return 0;
}