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 min(int a,int b){return a < b ? a : b;}

int main()
{
    int n,v[100000],num[100005][2] = {0},i,j,max[2][2] = {0};

    scanf("%d",&n);
    for (i = 0; i < n; i++)
    {
        scanf("%d",&v[i]);
        num[v[i]][i % 2]++;
    }

    for (i = 0; i < 2; i++)
    {
        for (j = 0; j <= 100000; j++)
        {
            if(num[max[i][0]][i] < num[j][i])
            {
                max[i][1] = max[i][0];
                max[i][0] = j;
            }
            else if (num[max[i][1]][i] < num[j][i])
            {
                max[i][1] = j;
            }
        }
    }
    if(max[0][0] != max[1][0])
    {
        printf("%d",n - num[max[0][0]][0] - num[max[1][0]][1]);
    }
    else if(max[0][1] == 0 && max[1][1] == 0)
    {
        printf("%d",n / 2);
    }
    else
    {
        printf("%d",min(n / 2 - max[0][1] + n / 2 - max[1][0] , n / 2 - max[1][1] + n / 2 - max[0][0]));
    }
    
}