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;
	scanf("%d",&n);
	int i,j;
	int asamecount,bsamecount;
	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);
	if(a[0]!=b[0]){
		int asame=1;
		int bsame=1;
		for(i=0;i<acount;i++){
			if(a[0]==a[i+1]){
				asame++;
			}
			if(b[0]==b[i+1]){
				bsame++;
			}
		}
		printf("%d",(acount+bcount)-(asame+bsame));
		return 0;
	}else{
		int asame=1;
		int bsame=1;
		for(i=0;i<acount;i++){
			if(a[0]==a[i+1]){
				asame++;
			}
			if(b[0]==b[i+1]){
				bsame++;
			}
		}
		if(asame>bsame){
			//bを二番目に多い数に置き換える
			//二番目に多い数の個数を求める
			i=0;
			while(b[i]=b[0]){
				i++;
			}
			if(i<acount){
				 bsamecount=1;
				
				j=i;
				while(b[i]==b[j]){
					j++;
					bsamecount++;
				}
			}else{
				bsamecount=0;
			}
			printf("%d",(acount*2)-asame-bsamecount);
		}else{
			//aを二番目に多い数に置き換える
			i=0;
			while(a[i]==a[0]){
				i++;
			}
			if(i<bcount){
				asamecount=1;
				j=i;
				while(a[i]=a[j]){
					j++;
					asamecount++;
				}
			}else{
				asamecount=0;
			}
			printf("%d",(acount*2)-asamecount-bsame);
		}
	}
	return 0;
}