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.*;

class Main{
  public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    int[] a=new int[n];
    for(int i=0;i<n;i++){
      a[i]=sc.nextInt();
    }
    int[] ki=new int[100001];
    int[] gu=new int[100001];
    for(int i=0;i<n;i+=2){
      gu[a[i]]++;
    }
    for(int i=1;i<n;i+=2){
      ki[a[i]]++;
    }
    int kif=0;
    int kis=0;
    int guf=0;
    int gus=0;
    for(int i=0;i<=100000;i++){
      if(ki[i]>ki[kif]){
        kis=kif;
        kif=i;
      }else if(ki[i]>ki[kis]){
        kis=i;
      }
    }
    for(int i=0;i<=100000;i++){
      if(gu[i]>gu[kif]){
        gus=guf;
        guf=i;
      }else if(gu[i]>gu[gus]){
        gus=i;
      }
    }
    if(kif!=guf){
      System.out.println(n-ki[kif]-gu[guf]);
    }else{
      System.out.println(
        Math.min(
          n-ki[kif]-gu[gus],
          n-ki[kis]-gu[guf]
        )
      );
    }
  }
}