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;

class Main{

	static long ans = 0;
	static int N;
	static long M;

	public static void main(String[] args) {



		Scanner sc = new Scanner(System.in);			//文字の入力

		int n = sc.nextInt();
		int[] v = new int[n];
		int[] odd= new int[n/2];
		int[] even= new int[n/2];


		int j = 0;
		for(int i=0;i < n;i++){
			if(i%2==1){
				odd[j] = sc.nextInt();
				j++;
			}else{
				even[j] = sc.nextInt();;
			}
		}


		boolean flag = false;				//全部一緒か見る


		for(int i=0;i < n/2-1;i++){
			if(odd[0] != even[0] || odd[i] != odd[i+1] || even[i] != even[i+1]){
				flag = true;
			}
		}
		if(!flag){
			System.out.println(odd.length);
			return;
		}
		j=0;


//
//		for(int i= 0;i < n/2;i++){
//			System.out.println(odd[i]);
//		}

		int[] odddata = new int[100001];
		for(int i = 0;i < n/2;i++){			//各数をカウント
			odddata[odd[i]]++;
		}
		int maxodd = 0;
		int firstodd = 0;
		int secondodd = odd[0];

		for(int i = 1;i < 100001;i++){			//最頻数の数値をえる
			if(maxodd<=odddata[i]){
				secondodd = firstodd;
				firstodd = i;
				maxodd = odddata[i];
			}
		}



		int[] evendata = new int[100001];
		for(int i = 0;i < n/2;i++){
			evendata[even[i]]++;
		}
		int maxeven= 0;
		int firsteven = 0;
		int secondeven = even[0];

		for(int i = 1;i < 100001;i++){
			if(maxeven<=evendata[i]){
				secondeven = firsteven;
				firsteven = i;
				maxeven = evendata[i];
			}
		}
	//	System.out.println("firsteven "+firsteven);

		int ans =0 ;
		j = 0;
		for(int i= 0;i < n;i++){
			if(i%2==1){
				if(odd[j] != firstodd){
					ans++;
				}
				j++;
			}else{
				if(even[j] != firsteven){
					ans++;
				}

			}
		}


		j = 0;
		int count = 0;
		for(int i= 0;i < n;i++){
			if(i%2==1){
				if(odd[j] != firstodd){
					count++;
				}
				j++;
			}else{
				if(even[j] != secondeven){
					count++;
				}

			}
		}

		j = 0;
		int subcount = 0;
		for(int i= 0;i < n;i++){
			if(i%2==1){
				if(odd[j] != secondodd){
					subcount++;
				}
				j++;
			}else{
				if(even[j] != firsteven){
					subcount++;
				}

			}
		}

		System.out.println(Math.min(count, Math.min(ans, subcount)));








	}


}