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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int a[]=new int[N/2];
		int b[]=new int[N/2];
		int d[]=new int[100001]; 
		int f[]=new int[100001]; 
		int l=N/2;
		for(int i=0;i<N/2;i++) {
			a[i]=in.nextInt();
			b[i]=in.nextInt();
		}
		
		for (int i = 0; i < l; i++) {
			d[a[i]]++;				// 頻度を数える：a[i]の値に該当する頻度値を1増やす
		}
		int max=1;
		int maxse=1;
		for (int i = 1; i < 100001; i++) {
			if (d[i] > d[max]) {
				maxse=max;
				max = i;			// 最大頻度を求める
			}
		}
		for (int i = 1; i < 100001; i++) {
			if (d[i] > d[maxse]&&!(i==max)) {
				maxse = i;			// 最大頻度を求める
			}
		}
		
		for (int i = 0; i < l; i++) {
			f[b[i]]++;				// 頻度を数える：a[i]の値に該当する頻度値を1増やす
		}
		
		int max2=1;
		int maxse2=1;
		for (int i = 1; i < 100001; i++) {
			if (f[i] > f[max2]) {
				maxse=max2;
				max2 = i;			// 最大頻度を求める
			}
		}
		for (int i = 1; i < 100001; i++) {
			if (f[i] > f[maxse2]&&!(i==max2)) {
				maxse2 = i;			// 最大頻度を求める
			}
		}
		
	    if(l-d[max]==0&&l-f[max2]==0&&b[0]==a[0]) {
	    	System.out.println(l);
	    }else if(max==max2){
	    	System.out.println(Math.min(2*l-d[max]-f[maxse2],2*l-d[maxse]-f[max2]));
	    }else {
	    	System.out.println(l-d[max]+(l-f[max2]));	    	
	    }

	}
}