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>

#define NUM  100000

int main()
{
    int input_len, output;
    int input;
    int a1_input_cnt[NUM];
    int a2_input_cnt[NUM];
    int a1_1st, a1_2nd, a1_val_1st, a1_val_2nd;
    int a2_1st, a2_2nd, a2_val_1st, a2_val_2nd;
    int num;
 
    for(int i= 0; i < NUM; i++)
    {
        a1_input_cnt[i] = 0;
        a2_input_cnt[i] = 0;
    }
    a1_1st = 0;
    a2_1st = 0;   
    a1_2nd = 0;
    a2_2nd = 0;
    a1_val_1st = 0;
    a1_val_2nd = 0;
    a2_val_1st = 0;
    a2_val_2nd = 0;
    
    //長さの取得
    scanf("%d", &input_len);
    
    if(input_len == 2)
    {
        printf("0");
        return (0);
    }
    
    for(int i = 0; i < input_len; i++)
    {
        scanf("%d", &input);
        num = i % 2;
        if(num == 0)
        {
            a1_input_cnt[input]++;
        }
        else
        {
            a2_input_cnt[input]++;
        }
            
    }

    
    for(int i = 0; i < NUM; i++)
    {
        //a1
        if(a1_input_cnt[i] != 0)
        {
            if(a1_1st < a1_input_cnt[i])
            {
                a1_2nd = a1_1st;
                a1_val_2nd = a1_val_1st;
                
                a1_1st = a1_input_cnt[i];
                a1_val_1st = i;
            }
            else if(a1_2nd <  a1_input_cnt[i])
            {
                a1_2nd = a1_input_cnt[i];
                a1_val_2nd = i;
            }
        }
        
        
        
        //a2
        if(a2_input_cnt[i] != 0)
        {
            if(a2_1st < a2_input_cnt[i])
            {
                a2_2nd = a2_1st;
                a2_val_2nd = a2_val_1st;
                
                a2_1st = a2_input_cnt[i];
                a2_val_1st = i;
            }
            else if(a2_2nd <  a2_input_cnt[i])
            {
                a2_2nd = a2_input_cnt[i];
                a2_val_2nd = i;
            }
        }      

    }
    
    if(a1_val_1st == a2_val_1st)
    {
        if(a1_1st >= a2_1st)
        {
            output = input_len - ( a1_1st + a2_2nd );
        }
        else
        {
            output = input_len - ( a1_2nd + a2_1st );
        }
    }
    else
    {
         output = input_len - ( a1_1st + a2_1st );
    }
    
    // 出力
    printf("%d\n", output);
    return 0;
}