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: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] O = new int[N/2];
        int[] E = new int[N/2];
        for(int i = 0; i < N; i++){
            if(i % 2 == 0){
                O[i/2] = sc.nextInt();
            }else{
                E[i/2] = sc.nextInt();
            }
        }
        sc.close();
        Arrays.sort(O);
        Arrays.sort(E);
        int[][] oMax = new int[N/2][2];
        int t = 0;
        int tCount = 0;
        int idx = 0;
        for(int j = 0; j < N/2; j++){
            if(t == 0){
                t = O[j];
                tCount = 1;
            }else if(O[j] != t){
                oMax[idx][0] = tCount;
                oMax[idx][1] = t;
                idx++;
                t = O[j];
                tCount = 1;
            }else{
                tCount++;
            }
        }
        oMax[idx][0] = tCount;
        oMax[idx][1] = t;

        int[][] eMax = new int[N/2][2];
        t = 0;
        tCount = 0;
        idx = 0;
        for(int j = 0; j < N/2; j++){
            if(t == 0){
                t = E[j];
                tCount = 1;
            }else if(E[j] != t){
                eMax[idx][0] = tCount;
                eMax[idx][1] = t;
                idx++;
                t = E[j];
                tCount = 1;
            }else{
                tCount++;
            }
        }
        eMax[idx][0] = tCount;
        eMax[idx][1] = t;
        Arrays.sort(oMax, (x, y) -> y[0]-x[0]);
        Arrays.sort(eMax, (x, y) -> y[0]-x[0]);

        if(oMax[0][1] != eMax[0][1]){ System.out.println(N - oMax[0][0] - eMax[0][0]);}
        else if(oMax[0][0] + eMax[1][0] >= oMax[1][0] + eMax[0][0]){ System.out.println(N - oMax[0][0] - eMax[1][0]);}
        else{ System.out.println(N - oMax[1][0] - eMax[0][0]);}

    }

}