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: Accepted)

import java.util.*;
 
public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in); 

        int n = sc.nextInt();
        int[] v = new int[n];
        for(int i = 0; i < n; i++){
            v[i] = sc.nextInt();
        }
        sc.close();
        int[] even = new int[100001];
        int[] odd = new int[100001];
        for(int i = 0; i < n; i=i+2){
            even[v[i]]++;
            odd[v[i+1]]++;
        }
        if(max_index(even) != max_index(odd)){
            System.out.println((n/2 - even[max_index(even)]) + (n/2 - odd[max_index(odd)]));
        } else{
            if((n/2 - even[max_index(even)]) + (n/2 - second(odd)) > (n/2 - second(even)) + (n/2 - odd[max_index(odd)])){
                System.out.println((n/2 - second(even)) + (n/2 - odd[max_index(odd)]));
            } else{
                System.out.println((n/2 - even[max_index(even)]) + (n/2 - second(odd)));
            }
        }
       
    }
    static int max_index(int[] a){
        int index = 0;
        int max = a[0];
        for(int i = 1; i < a.length; i++){
            if(a[i] > max){
                max = a[i];
                index = i;
            }
        }
        return index;
    }
    
    static int second(int[] a){
        int max1 = 0;
        int max2 = 0;
        for(int i = 0; i < a.length; i++){
            if(a[i] > max1){
                max2 = max1;
                max1 = a[i];
            }else if(a[i] > max2){
                max2 = a[i];
            }
        }
        return max2;
    }
}