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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner scan = new java.util.Scanner(System.in);
        int n = scan.nextInt();
        int[] a = new int[n / 2];
        int[] b = new int[n / 2];
        int ma = 0;
        int mb = 0;
        
        for(int i = 0; i < n; i++){
            if(i % 2 == 0){
                a[i / 2] = scan.nextInt();
            }else{
                b[i / 2] = scan.nextInt();
            }
        }
        Arrays.sort(a);
        Arrays.sort(b);
        
        for(int i = 0; i < n / 2 - 1; i++){
            if(a[i] != a[i + 1]){
                ma++;
            }else if(b[i] != b[i + 1]){
                mb++;
            }
        }
        if(Arrays.equals(a, b)){
            System.out.println(n / 2);
        }else{
            System.out.println(ma + mb);
        }
    }
}