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: Accepted)

#include <stdio.h>
int main(void)
{
    int n, a[10*10*10*10*10*10], ans, cnteven[10*10*10*10*10*10], i;
    int cntodd[10*10*10*10*10*10], maxeven=0, maxodd=0, max=0,
        evenidx, oddidx;
    int maxodd2=0, maxeven2=0;
    scanf("%d", &n);
    for(i=0;i<n;i++){
        scanf("%d", &a[i]);
        if(a[i]>max) max=a[i];
    }

    for(i=0;i<n;i++){
        if(i%2==0) cnteven[a[i]]++;
        else cntodd[a[i]]++;
    }

    for(i=0;i<=max;i++){
        if(maxeven<cnteven[i]){
            maxeven2=maxeven;
            maxeven=cnteven[i];
            evenidx=i;
        }
        else if(maxeven2<cnteven[i]){
            maxeven2=cnteven[i];
        }
        if(maxodd<cntodd[i]){
            maxodd2=maxodd;
            maxodd=cntodd[i];
            oddidx=i;
        }
        else if(maxodd2<cntodd[i]){
            maxodd2=cntodd[i];
        }
    }

    if(oddidx != evenidx) ans=n-maxeven-maxodd;
    else if(n-maxodd-maxeven2<n-maxodd2-maxeven) ans=n-maxodd-maxeven2;
    else ans=n-maxodd2-maxeven;
    
    printf("%d\n", ans);
    return 0;
}