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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n / 2];
		int[] b = new int[n / 2];

		//配列vの奇数行・偶数行を2つの配列a,bに分ける。(O(N))
		for (int i = 0; i < n / 2; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
		}

		//a,bそれぞれについて、値が出現した回数を配列s,tでカウントする。(O(2 * 10^5))
		int[] s = new int[100_005];
		int[] t = new int[100_005];

		for (int i = 0; i < n / 2; i++) {
			s[a[i]]++;
			t[b[i]]++;
		}


		//ans = n - (sの最大値) - (tの最大値) (O(2 * 10^5))
		int ms = 0;
	    int mt = 0;
	    int sms = 0;
		int smt = 0;

	    int ma = 0;
	    int mb = 0;
	    int sma = 1;
	    int smb = 1;
		for (int i = 0 ; i < 100_005; i++) {
			if (s[i] >= ms) {
				sms = ms;
				ms = s[i];

				sma = ma;
				ma = i;
			}
			if (t[i] >= mt) {
				smt = mt;
				mt = t[i];

				smb = mb;
				mb = i;
			}
		}

		//ms,mtをとるiの値が一致するとき、sとtのセカンドmaxをms,mtとして、
		//最小化のansを求める。
		int ans = 0;
		if (ma != mb) {
			ans = n - ms - mt;
		} else {
			ans = Math.min(n - ms - smt, n - sms - mt);
		}

		System.out.println(ans);

	}

}