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) throws Exception{
		Scanner stdIn=new Scanner(System.in);
		int n=stdIn.nextInt();
		int v=0;
		int v1=0;
		int v2=0;
		int odd[][]=new int[100001][2];
		int even[][]=new int[100001][2];
		int z=0,y=0,x=0,cou=0;
		while(z<n){
			v=stdIn.nextInt();
			if(z==0)
				v1=v;
			if(z==1)
				v2=v;
			if(z%2==0){
				odd[v][0]++;
				odd[v][1]=v;
				x++;
			}
			else{
				even[v][0]++;
				even[v][1]=v;
				y++;
			}
			z++;
		}z=0;
		Arrays.sort(odd, (a, b) -> Long.compare(a[0], b[0]));
		Arrays.sort(even, (a, b) -> Long.compare(a[0], b[0]));
		if(x!=odd[100000][0]){
			cou+=x-odd[100000][0];
		}
		if(y!=even[100000][0]){
			cou+=y-even[100000][0];
		}
		if(x!=odd[100000][0]&&y!=even[100000][0]){
			if(even[100000][1]==odd[100000][1])
				if(odd[100000][0]-odd[99999][0]>=even[100000][0]-even[99999][0])
					cou+=odd[100000][0]-odd[99999][0];
				else
					cou+=even[100000][0]-even[99999][0];
		}
		if(x==odd[100000][0]&&y==even[100000][0]){
			if(v1==v2)
				cou+=y;
		}
		System.out.println(cou);
	}
}