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 (2.3.3) (judge verdict: Wrong Answer)

n = gets.chomp.to_i

odd_max_x = -1
odd_second_x = -1
even_max_x = -1
even_second_x = -1
h_odd = { -1 => 0 }
h_even = { -1 => 0 }
a, b = gets.chomp.split(' ').map(&:to_i).partition.with_index(0) do |x, i|
  even =  i.zero? || i.even?
  h = even ? h_even : h_odd
  if h.key?(x)
    h[x] += 1
  else
    h[x] = 1
  end

  if even
    if h[x] > h[even_max_x]
      even_second_x = even_max_x
      even_max_x = x
    elsif h[x] > h[even_second_x]
      even_second_x = x
    end
  else
    if h[x] > h[odd_max_x]
      odd_second_x = odd_max_x
      odd_max_x = x
    elsif h[x] > h[odd_second_x]
      odd_second_x = x
    end
  end
  even
end

if even_max_x != odd_max_x
  puts n - h_even[even_max_x] - h_odd[odd_max_x]
  exit
end

use_max_even = h_even[even_max_x] + h_odd[odd_second_x]
use_max_odd = h_odd[odd_max_x] + h_even[even_second_x]

def single?(h)
  h.keys.length == 2
end

if single?(h_even) && single?(h_odd)
  puts n / 2
elsif single?(h_even) && !single?(h_odd)
  puts (n / 2) - h_odd[odd_second_x]
elsif !single?(h_even) && single?(h_odd)
  puts (n / 2) - h_even[even_second_x]
else
  puts n - [use_max_even, use_max_odd].min
end