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.*;
import java.math.*;
import java.util.stream.Collectors;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		int n = sc.nextInt();
		List<Integer> oddList = new ArrayList<>();
		List<Integer> evenList = new ArrayList<>();

		for(int i=0;i<n;i=i+2){
			oddList.add(sc.nextInt());
			evenList.add(sc.nextInt());
		}
		Collections.sort(oddList);
		Collections.sort(evenList);

		Map<Integer, Long> oddMap = new HashMap<>();
		Map<Integer, Long> evenMap = new HashMap<>();
		//それぞれで一番多い数と二番目を特定しておく
		oddMap = oddList.stream().collect(
                Collectors.groupingBy(x -> x, Collectors.counting())
        );
        evenMap = evenList.stream().collect(
                Collectors.groupingBy(x -> x, Collectors.counting())
        );

		long maxOdd =0;
		long maxEven =0;
		long secondOdd =0;
		long secondEven =0;
		// long thirdOdd =0;
		// long thirdEven =0;
		int maxOddValue =0;
		int maxEvenValue =0;
		int secondOddValue =0;
		int secondEvenValue =0;
		// int thirdOddValue =0;
		// int thirdEvenValue =0;
		for(Iterator<Integer> iterator = oddMap.keySet().iterator(); iterator.hasNext(); ) {
		    Integer key = iterator.next();
		    // System.out.println(key + " => " + oddMap.get(key));
		    if(maxOdd<=oddMap.get(key)){
		    	// thirdOdd =secondOdd;
		    	// thirdOddValue =secondOddValue;
		    	secondOdd =maxOdd;
		    	secondOddValue =maxOddValue;

		    	maxOdd=oddMap.get(key);
		    	maxOddValue = key;
		    }
		}

		for(Iterator<Integer> iterator = evenMap.keySet().iterator(); iterator.hasNext(); ) {
		    Integer key = iterator.next();
		    // System.out.println(key + " => " + evenMap.get(key));
		    if(maxEven<=evenMap.get(key)){
		    	// thirdEven =secondEven;
		    	// thirdEvenValue =secondEvenValue;
		    	secondEven =maxEven;
		    	secondEvenValue =maxEvenValue;

		    	maxEven=evenMap.get(key);
		    	maxEvenValue = key;
		    }
		}

		//２つが一緒の場合はどちらかをずらす。そうでなければその数のまま
		if(maxEvenValue!=maxOddValue){
			System.out.println(n-maxEven-maxOdd);
		}else{
			//decide change is which
			if(maxOdd>maxEven){
				// System.out.println("oddの方が多い");
				System.out.println(n-secondEven- maxOdd);
			}else if (maxOdd<maxEven){
				// System.out.println("evenの方が多い");
				System.out.println(n-maxEven- secondOdd);
			}else{
				// System.out.println("同数だから2つ目のやつの多い方を採用する");
				System.out.println(n-maxEven- Math.max(secondOdd,secondEven));
			}
		}
		// System.out.println("log");
		// System.out.println(oddMap);
		// System.out.println(evenMap);

	}

}