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: Accepted)

import java.util.*;

public class Main {

    public static void main(String args[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        Integer[] a = new Integer[n];
        Arrays.setAll(a, i -> Integer.parseInt(sc.next()));
        sc.close();

        // 主処理
        // 値の出現回数をカウント
        Integer[] oddIndexCount = new Integer[100001];
        Arrays.fill(oddIndexCount, 0);
        for (int i = 0; i < n; i += 2) {
            oddIndexCount[a[i]]++;
        }
        Integer[] evenIndexCount = new Integer[100001];
        Arrays.fill(evenIndexCount, 0);
        for (int i = 1; i < n; i += 2) {
            evenIndexCount[a[i]]++;
        }

        // 最大と2番目の出現回数を記録
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        pq.addAll(Arrays.asList(oddIndexCount));
        int oddIndexCountFirst = pq.poll();
        int oddIndexCountSecond = pq.poll();
        pq = new PriorityQueue<>(Collections.reverseOrder());
        pq.addAll(Arrays.asList(evenIndexCount));
        int evenIndexCountFirst = pq.poll();
        int evenIndexCountSecond = pq.poll();

        // 最大個数の値を取得
        int oddIndexNumber = 0;
        for (int i = 1; i < oddIndexCount.length; i++) {
            if (oddIndexCount[i] == oddIndexCountFirst) {
                oddIndexNumber = i;
                break;
            }
        }
        int evenIndexNumber = 0;
        for (int i = evenIndexCount.length - 1; 1 <= i; i--) {
            if (evenIndexCount[i] == evenIndexCountFirst) {
                evenIndexNumber = i;
                break;
            }
        }

        int result = n;
        if (oddIndexNumber == evenIndexNumber) {
            result -= Math.max(oddIndexCountFirst + evenIndexCountSecond, oddIndexCountSecond + evenIndexCountFirst);
        } else {
            result -= (oddIndexCountFirst + evenIndexCountFirst);
        }

        // 出力
        System.out.println(result);
    }
}