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 comp( const void * a , const void * b ) {
	if(((int *)a)[1] < ((int *)b)[1]) {
		return 1;
	}
	else
	if(((int *)a)[1] == ((int *)b)[1]) {
		return 0;
	}
	return -1;
}

int main()
{
	int NMAX = 100000;
	int n;
	int v_odd[NMAX][2];
	int v_even[NMAX][2];

	for (int i = 0; i < NMAX; i ++) {
		v_odd[i][0] = i+1;
		v_odd[i][1] = 0;
		v_even[i][0] = i+1;
		v_even[i][1] = 0;
	}

	scanf("%d", &n);
	for (int i = 0; i < (n / 2); i ++) {
		int t;
		scanf("%d", &t);
		v_odd[t-1][1] ++;
		scanf("%d", &t);
		v_even[t-1][1] ++;
	}

	qsort( (void*)v_odd, NMAX, sizeof(v_odd[0]), comp );
	qsort( (void*)v_even, NMAX, sizeof(v_even[0]), comp );

	/*
	printf ("-- dump --\n");
	for (int i = 0; i < NMAX; i ++) {
		printf("(%d %d)", v_odd[i][0], v_odd[i][1]);
	}
	printf ("\n");
	for (int i = 0; i < NMAX; i ++) {
		printf("(%d %d)", v_even[i][0], v_even[i][1]);
	}
	*/

	if (v_odd[0][0] == v_even[0][0]) {
		if (v_odd[0][1] > v_even[0][1]) {
			printf("%d", (n/2 - v_odd[0][1]) + (n/2 - v_even[1][1]));
		} else {
			printf("%d", (n/2 - v_odd[1][1]) + (n/2 - v_even[0][1]));
		}
	} else {
		printf("%d", (n/2 - v_odd[0][1]) + (n/2 - v_even[0][1]));
	}
	return 0;
}