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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int n,v[100005],m[100005],m1[100005],p1,p2,vp1,vp2,n1,n2,vn1,vn2;

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&v[i]);
    for(int i=0;i<n;i+=2)
    {
        m[v[i]]++;
        if(m[v[i]]>p1)
        {
            if(vp1==v[i])
                p1=m[v[i]];
            else
            {
                p2=p1;
                vp2=vp1;
                p1=m[v[i]];
                vp1=v[i];
            }
        }
        else if(m[v[i]]>p2&&v[i]!=vp1)
        {
            p2=m[v[i]];
            vp2=v[i];
        }
    }
    for(int i=1;i<n;i+=2)
    {
        m1[v[i]]++;
        if(m1[v[i]]>n1)
        {
            if(vn1==v[i])
                n1=m1[v[i]];
            else
            {
                n2=n1;
                vn2=vn1;
                n1=m1[v[i]];
                vn1=v[i];
            }
        }
        else if(m1[v[i]]>n2&&v[i]!=vn1)
        {
            n2=m1[v[i]];
            vn2=v[i];
        }
    }
    if(vp1!=vn1)
    {
        printf("%d",n-p1-n1);
        return 0;
    }
    else
    {
        if(n1>p1)
        {
            printf("%d",n-n1-p2);
            return 0;
        }
        else
        {
            printf("%d",n-p1-n2);
            return 0;
        }
    }

    return 0;
}