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: Wrong Answer)

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
        int[] oddIndexNumberCount = new int[100001];
        for (int i = 0; i < n; i += 2) {
            oddIndexNumberCount[a[i]]++;
        }
        int oddIndexNumberCountFirst = 0;
        int oddIndexNumberCountSecond = 0;
        for (int i = 0; i < oddIndexNumberCount.length; i++) {
            if (oddIndexNumberCountFirst <= oddIndexNumberCount[i]) {
                oddIndexNumberCountSecond = oddIndexNumberCountFirst;
                oddIndexNumberCountFirst = oddIndexNumberCount[i];
            }
        }
        int oddIndexNumber = 0;
        for (int i = 0; i < oddIndexNumberCount.length; i++) {
            if (oddIndexNumberCount[i] == oddIndexNumberCountFirst) {
                oddIndexNumber = i;
            }
        }

        int[] evenIndexNumberCount = new int[100001];
        for (int i = 1; i < n; i += 2) {
            evenIndexNumberCount[a[i]]++;
        }
        int evenIndexNumberCountFirst = 0;
        int evenIndexNumberCountSecond = 0;
        for (int i = 0; i < evenIndexNumberCount.length; i++) {
            if (evenIndexNumberCountFirst <= evenIndexNumberCount[i]) {
                evenIndexNumberCountSecond = evenIndexNumberCountFirst;
                evenIndexNumberCountFirst = evenIndexNumberCount[i];
            }
        }
        int evenIndexNumber = 0;
        for (int i = evenIndexNumberCount.length - 1; 0 <= i; i--) {
            if (evenIndexNumberCount[i] == evenIndexNumberCountFirst) {
                evenIndexNumber = i;
            }
        }

        int result = 0;
        if (oddIndexNumber == evenIndexNumber) {
            result += Math.min(n - oddIndexNumberCountFirst - evenIndexNumberCountSecond,
                    n - oddIndexNumberCountSecond - evenIndexNumberCountFirst);

        } else {
            result += n - oddIndexNumberCountFirst - evenIndexNumberCountFirst;
        }

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