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 Ruby (2.7.1) (judge verdict: Wrong Answer)

require 'benchmark'

gets.to_i
inputs = gets.split(" ").map(&:to_i)
odd = inputs.each_slice(2).map(&:first)
even = inputs.each_slice(2).map(&:last)
# odd_max = odd.max_by{|v| odd.count(odd)}
# even_max = even.max_by{|v| even.count(even)}
odd_freq_ary = odd.group_by { |e| e }.sort_by { |e, v| -v.size }.map(&:first)
odd_max = odd_freq_ary.first
even_freq_ary = even.group_by { |e| e }.sort_by { |e, v| -v.size }.map(&:first)
even_max = even_freq_ary.first
if odd_max == even_max
  # 2番目に頻度多いのはどっちか
  odd_second_freq_count = odd.count(odd_freq_ary[1])
  even_second_freq_count = even.count(even_freq_ary[1])
  # どっちも全部一緒だったら片方全部書き換え
  if odd_second_freq_count.nil? && even_second_freq_count.nil?
    puts odd.count
  elsif odd_second_freq_count == even_second_freq_count
    num = 0
    num += (odd - [odd_freq_ary[1]]).size
    num += (even - [even_max]).size
    puts num
  elsif odd_second_freq_count > even_second_freq_count
    num = 0
    num += (odd - [odd_freq_ary[1]]).size
    num += (even - [even_max]).size
    puts num
  else
    num = 0
    num += (odd - [odd_max]).size
    num += (even - [even_freq_ary[1]]).size
    puts num
  end
else
  num = 0
  num += (odd - [odd.max]).size
  num += (even - [even.max]).size
  puts num
end