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>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <limits.h>
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r);
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r);
#define TYPE long long
#define MEMSET(a) memset(a,0,sizeof(a))
const int mod=(int)1e09+7;

int in(void){
    int i;scanf("%d",&i);
    return i;
}
double din(void){
    double i;scanf("%lf",&i);
    return i;
}
void chin(char s[]){
    scanf("%s",s);
}
void print(int a){
    printf("%d\n",a);
}
void llprint(long long a){
    printf("%lld\n",a);
}
void dprint(double a){
    printf("%.10f\n",a);
}
void print2(int a,int b){
    printf("%d %d\n",a,b);
}
double dmax(double a,double b){
    return a>b?a:b;
}

int char_cmp(const void *a,const void *b){
    return strcmp((char *)a,(char *)b);
}
int char_cmp_r(const void *a,const void *b){
    return strcmp((char *)b,(char *)a);
}
void swap(int *a,int *b){
    int t=*a;
    *a=*b;
    *b=t;
}

int ruizyou(int a, int b){
    int i, num=1;
    rep(i, b){
        num = num*a;
    }
    return num;
}

int main(void){
    int i, j, n;
    n = in();
    int v[n];
    rep(i, n){
        v[i] = in();
    }
    int ans;
    int fre_1, fre_2;
    int a[100000], b[100000];             //a[i]:iの出現回数(偶), b[i]:iの出現回数(奇)
    rep(i, 100000){
        a[i] = 0;                         //初期化
        b[i] = 0;
        rep(j, n/2){
            if(i==v[2*j]) a[i]++;         //数列のj番目と同じか判定
            if(i==v[2*j+1]) b[i]++;
        }
    }
    int max_a = 0;                     //max_a, max_bを同じ数字の数とする
    int max_a_1 = 0;                      //次に大きい数
    int max_b = 1;
    int max_b_1 = 0;
    rep(i, 100000){
        if(a[i]>=a[max_a]){
            max_a_1 = max_a;
            max_a = i;
        }
        if(b[i]>=b[max_b]){
            max_b_1 = max_b;
            max_b = i;
        }
    }
    if(max_a==max_b){
        if(a[max_a]+b[max_b_1]>a[max_a_1]+b[max_b]) max_b = max_b_1;
        else max_a = max_a_1;
    }
    ans = n - a[max_a] - b[max_b];
    print(ans);
    return 0;
}