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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int[] vo = new int[n / 2];
		int[] ve = new int[n / 2];

		for (int i = 0; i < n / 2; i++) {
			vo[i] = sc.nextInt();
			ve[i] = sc.nextInt();
		}

		Arrays.sort(vo);
		Arrays.sort(ve);
		int maxo = 1;
		int maxe = 1;
		int seco = 0;
		int sece = 0;
		int numo = 0;
		int nume = 0;

		int j = 0, k = 0;
		int ans = -1;

		while (j != n / 2 - 1) {
			int count = 1;
			while (vo[j] == vo[j + 1]) {
				count++;
				j++;
				if (j == n / 2 - 1) {
					break;
				}
			}
			if (j != n / 2 - 1) {
				j++;
			}
			if (maxo <= count) {
				seco = maxo;
				maxo = count;
				numo = vo[j - 1];
			}

		}

		while (k != n / 2 - 1) {
			int count = 1;
			while (ve[k] == ve[k + 1]) {
				count++;
				k++;
				if (k == n / 2 - 1) {
					break;
				}
			}
			if (k != n / 2 - 1) {
				k++;
			}
			if (maxe <= count) {
				sece = maxe;
				maxe = count;
				nume = ve[k - 1];
			}

		}

		if (maxo == n / 2 && maxe == n / 2) {
			if (vo[0] == ve[0]) {
				ans = n / 2;
			} else {
				ans = 0;
			}
		} else if (numo != nume) {
			ans = (n / 2 - maxo) + (n / 2 - maxe);
		} else {
			ans = Math.min((n / 2 - maxo) + (n / 2 - sece), (n / 2 - maxe) + (n / 2 - seco));
		}

//		System.out.println(maxo + "," + maxe + "," + seco + "," + sece + "," + numo + "," + nume);
		System.out.println(ans);

	}
}