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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

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

int main(void){
    int n;
    scanf("%d", &n);
    int v[n];
    for(int i = 0; i < n; i++){scanf("%d", &v[i]);}
    int w[n];
    for(int i = 0; i < n; i++){w[i] = v[i];}
    qsort(w, n, sizeof(int), comp);
    //文字の種類を数える。文字の種類はcount+1個
    int idx = 0;
    int tmp = w[0];
    int count = 0;
    int d[n];
    d[0] = tmp;
    while(idx < n){
        if(w[idx] != tmp){
            tmp = w[idx];
            count++;
            d[count] = tmp;
        }
        idx++;
    }
    count++;
    if(count == 1){printf("%d\n", n/2); return 0;}
    //ある数字を奇数、偶数番目にするときの必要な変更回数を求める
    int pointEven, pointOdd;
    int minOdd = n+1;
    int minEven = n+1;
    int p[count][2];
    for(int i = 0; i < count; i++){
        int target = d[i];
        pointEven = 0;
        pointOdd= 0;
        for(int j = 1; j < n; j+=2){
            if(v[j] != target){pointEven++;}
        }
        p[i][0] = pointEven;
        for(int j = 0; j < n; j+=2){
            if(v[j] != target){pointOdd++;}
        }
        p[i][1] = pointOdd;
    }
    //どう考えても(10^5)^2で時間オーバーだけどこれしか思いつかなかったので。
    int min = n;
    for(int i = 0; i < count; i++){
        for(int j = 0; j < count; j++){
            if(i == j){continue;}
            int sum = p[i][0] + p[j][1];
            if(min > sum){min = sum;}
            sum = 0;
        }
    }
    printf("%d\n", min);
    return 0;
}