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>

int main(void)
{
    int n,v[100000],t1[100000]={0},t2[100000]={0},i,a=0;
    scanf("%d",&n);
    for(i=0;i<n;++i){
        scanf("%d",&v[i]);
        if(i%2==0)++t1[v[i]-1];
        else ++t2[v[i]-1];
    }

    //最頻値を調べる
    int m1=0,m2=0;
    for(i=0;i<100000;++i){
        if(t1[i]>m1)m1=i+1;
        if(t2[i]>m2)m2=i+1;
    }

    //m1, m2を出力
    //printf("%d %d\n",m1,m2);

    if(m1==m2){

        //最頻値の個数を調べる
        int cnt1=0,cnt2=0;
        for(i=0;i<n;++i){
            if(i%2==0){
                if(v[i]==m1)++cnt1;
            }
            else{
                if(v[i]==m2)++cnt2;
            }
        }

        //cnt1, cnt2出力
        //printf("%d %d\n",cnt1,cnt2);

        //第2の最頻値を調べる
        int sec1=0,sec2=0;
        for(i=0;i<100000;++i){
            if(t1[i]>sec1&&m1!=i+1)sec1=i+1;
            if(t2[i]>sec2&&m1!=i+1)sec2=i+1;
        }

        //sec1, sec2を出力
        //printf("%d %d\n",sec1,sec2);

        if(cnt1>cnt2){
            for(i=0;i<n;++i){
                if(i%2==0){
                    if(v[i]!=m1)++a;
                }
                else{
                    if(v[i]!=sec2)++a;
                }
            }
        }

        else{
            for(i=0;i<n;++i){
                if(i%2==0){
                    if(v[i]!=sec1)++a;
                }
                else{
                    if(v[i]!=m2)++a;
                }
            }
        }

    }
    
    else{

        for(i=0;i<n;++i){
            if(i%2==0){
                if(v[i]!=m1)++a;
            }
            else{
                if(v[i]!=m2)++a;
            }
        }

    }

    printf("%d",a);
    return 0;
}