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 main(void) {
	int n;
 
  	if(scanf("%d", &n) == 1);
 
	int guu[100000];
	int ki[100000];

	int gucount = 0;
	int baka = 0;
	int bakavalue = 0;
	int guuindex = 0;

	int kicount = 0;
	int aho = 0;
	int ahovalue = 0;
	int kiindex = 0;

	int input;
	int num;
	int output = 0;

	if(n < 2 || n > 100000 || n % 2 != 0) {
		exit(0);
	}

	if(n == 2) {
        printf("0");
        return 0;
    }

    for(int i = 0; i < 100000; i++) {
        guu[i] = 0;
        ki[i] = 0;
    }

    for(int i = 0; i < n; i++) {
    	if(scanf("%d", &input) == 1);
        num = i % 2;
        if(num == 0) {
            guu[input]++;
        } else {
            ki[input]++;
        }
    }

    for(int i = 0; i < 100000; i++) {
        if(guu[i] != 0) {
            if(gucount < guu[i]) {
            	// 2番目のcount(偶数)
                baka = gucount;
                // 2番目のindex(偶数)
                bakavalue = guuindex;
                // 1番目のcount(偶数)
                gucount = guu[i];
                // 1番目のindex(偶数)
                guuindex = i;
            } else if(baka < guu[i]) {
            	// 2番目のcount(偶数)
                baka = guu[i];
                // 2番目のindex(偶数)
                bakavalue = i;
            }
        }
        
        if(ki[i] != 0) {
            if(kicount < ki[i]) {
                aho = kicount;
                ahovalue = kiindex;
                kicount = ki[i];
                kiindex = i;
            } else if(aho < ki[i]) {
                aho = ki[i];
                ahovalue = i;
            }
        }
    }

    if(guuindex == kiindex) {
    	if(gucount >= kicount) {
    		output = (n / 2 - gucount) + (n / 2 - aho);
    	} else {
    		output = (n / 2 - baka) + (n / 2 - kicount);
    	}
    } else {
    	output = (n / 2 - gucount) + (n / 2 - kicount);
    }

	printf("%d\n", output);
}