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) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner scanner = new Scanner(System.in);
		
		int counter = scanner.nextInt();
		int[] odd = new int[100001];
		int[] even = new int[100001];
		
		boolean bool = true;
		for(int i = 0;i<counter;i++) {
			if(bool) odd[scanner.nextInt()]++;
			else even[scanner.nextInt()]++;
			bool = !bool;
		}
		
		int a=0,b=0,c=0,d=0;
		
		for(int i = 0;i<100001;i++) {
			if(odd[i] > a) {
				a = odd[i];
				b = i;
			}
			if(even[i] > c) {
				c = even[i];
				d = i;
			}
		}
		
		bool = b == d;
		
		Arrays.sort(odd);
		Arrays.sort(even);
		
		if(!bool) {
			System.out.println(counter - odd[100000] - even[100000]);
		}else {
			if(odd[100000] > even[100000]) {
				System.out.println(counter - odd[100000] - even[99999]);
			}else {
				System.out.println(counter - odd[99999] - even[100000]);
			}
		}
	}
}