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 compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int main(void){
	int n,ans=0;
	int ans2=0;
	scanf("%d",&n);
	int i,j;
	int v[n];
	for(i=0;i<n;i++){
		scanf("%d",&v[i]);
	}
	int a[n];
	int b[n];
	int acount=0;
	int bcount=0;
	for(i=0;i<n;i++){
		if(i%2==0){
			a[acount]=v[i];
			acount++;
		}else{
			b[bcount]=v[i];
			bcount++;
		}
	}
	//後はa[]とb[]をソートして最も多い数に合わせれば良い
	qsort(a, n/2, sizeof(int), compare_int);
	qsort(b, n/2, sizeof(int), compare_int);
	//a[]で最も出現する数の個数をカウントする。
	int a1ban=0;
	int a2ban=0;
	int b1ban=0;
	int b2ban=0;
	i=0;
	while(a[0]==a[i]){
		i++;
		a1ban++;
	}
	if(i<n/2){
		while(a[a1ban]==a[i]){
			i++;
			a2ban++;
		}
	}
	//printf("a1=%d a2=%d\n",a1ban,a2ban);
	i=0;
	while(b[0]==b[i]){
		i++;
		b1ban++;
	}
	if(i<n/2){
		while(b[b1ban]==b[i]){
			i++;
			b2ban++;
		}
	}
	//printf("b1=%d b2=%d\n",b1ban,b2ban);
	if(a[0]!=b[0]){
		ans=n-a1ban-b1ban;
	}else{
		ans=n-a1ban-b2ban;
		ans2=n-a2ban-b1ban;
		if(ans>ans2){
			ans=ans2;
		}
	}
	printf("%d",ans);
	return 0;
}