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>
#include<stdlib.h>
int asc(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}
int main()
{
    int n;
    scanf("%d", &n);
    int i;
    int v[100005];
    for (i = 0; i < n; i++)
        scanf("%d", &v[i]);
    int o[100005], e[100005];
    for (i = 0; i < n; i++)
        if (i % 2 == 0)
            e[i / 2] = v[i];
        else
            o[i / 2] = v[i];
    int m = n / 2;
    qsort(e, m, sizeof(int), asc);
    qsort(o, m, sizeof(int), asc);
    e[m] = o[m] = -1;
    int e1, ne1, o1, no1;
    ne1 = no1 = 0;
    int x = 1;
    for (i = 1; i <= m; i++)
        if (e[i] == e[i - 1])
            x++;
        else
        {
            if (x > ne1)
                ne1 = x;
            e1 = e[i - 1];
            x = 1;
        }
    x = 1;
    for (i = 1; i <= m; i++)
        if (o[i] == o[i - 1])
            x++;
        else
        {
            if (x > no1)
                no1 = x;
            o1 = o[i - 1];
            x = 1;
        }
    if (e1 != o1)
    {
        printf("%d\n", n - ne1 - no1);
        return 0;
    }
    int no2, ne2;
    no2 = ne2 = 0;
    if (e[0] == e1)
        x = 0;
    else
        x = 1;
    for (i = 1; i <= m; i++)
        if (e[i] == e[i - 1] && e[i] != e1)
            x++;
        else
        {
            if (x > ne2)
                ne2 = x;
            if (e[i] == e1)
                x = 0;
            else
                x = 1;
        }
    if (o[0] == o1)
        x = 0;
    else
        x = 1;
    for (i = 1; i <= m; i++)
        if (o[i] == o[i - 1] && o[i] != o1)
            x++;
        else
        {
            if (x > no2)
                no2 = x;
            if (o[i] == o1)
                x = 0;
            else
                x = 1;
        }
    if (no2 < ne2)
        ne1 = ne2;
    else
        no1 = no2;
    printf("%d\n", n - no1 - ne1);
    return 0;
}