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,c[100001],d[100001],max1[2],max2[2],ans,i,x,y,p=0,q=0;
    scanf("%i",&n);
    max1[0]=max1[1]=0;
    max2[0]=max2[1]=0;
    int a[n/2],b[n/2];
    for(i=0;i<=100000;i++){c[i]=0;d[i]=0;}
    for(i=0;i<n/2;i++)
    {
        scanf("%i %i",&a[i],&b[i]);
        c[a[i]]++;
        d[b[i]]++;
    }
    for(i=0;i<=100000;i++)
    {
        if(max1[0]<c[i]){max1[1]=max1[0];max1[0]=c[i];x=i;}
        if(max2[0]<d[i]){max2[1]=max2[0];max2[0]=d[i];y=i;}
    }
    for(i=0;i<=100000;i++)
    {
        if(max1[0]==c[i]){p++;}
        if(max2[0]==d[i]){q++;}
    }
    if(p==1&&q==1)
    {
        if(x!=y)ans=n-max1[0]-max2[0];
        else if(max1[1]<max2[1])ans=n-max1[0]-max2[1];
        else if(max2[1]<max1[1])ans=n-max2[0]-max1[1];
        else ans=n-max1[0];
    }
    else
    {
        if(x!=y)ans=n-max1[0]-max2[0];
        else ans=n-max1[0];
    }
    printf("%i",ans);
}