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>
#include <string.h>

int main()
{
    int N;
    scanf("%d", &N);

    int array[N];
     
    int M = 1e5 * 2;
    int EvenArr[M];
    int OddArr[M];

    memset(EvenArr, 0, sizeof(EvenArr[0]) * M);
    memset(OddArr, 0, sizeof(OddArr[0]) * M);

    for (int i = 0; i < N; i++)
    {
        scanf("%d", &array[i]);
        if (i % 2 == 0)
            OddArr[array[i]]++;
        else if (i % 2 != 0)
            EvenArr[array[i]]++;
    }
    int cnt = 0;
    for (int i = 0; i < N - 1; i++)
    {
        if (array[i] == array[i + 1])
            cnt++;
    }
    if (cnt == N - 1)
    {
        printf("%d", N / 2);
        return 0;
    }

    int count = 0;
    for (int i = 0; i < N - 2; i++)
    {
        if (i % 2 == 0)
        {
            if (array[i] != array[i + 2])
            {
                if (EvenArr[ array[i] ] > EvenArr[ array[i+2] ])

                {
                array[i] = array[i + 2];
                count++;
                }

                else 
                {
                array[i + 2] = array[i];
                count++;
                }
            }
        }

        else  
        {    
          if (array[i] != array[i + 2])
            {
                if (OddArr[ array[i] ] > OddArr[ array[i+2] ])

                {
                array[i] = array[i + 2];
                count++;
                }

                else 
                {
                array[i + 2] = array[i];
                count++;
                }

            }
        }
    }

    printf("%d", count);

    return 0;
}