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: Runtime Error)

 = gets.chomp.to_i
a = gets.chomp.split(" ").map(&:to_i)
a_rev = a.reverse

if a == a_rev
  print("#{a.size()/2}")
  exit
end

b1 = Array.new(0)
b1.fill(0,a.size()){|idx| 
  if idx % 2 == 0
    a[idx]
  else 
    a_rev[idx]
  end
  }
b2 = Array.new(0)
b2.fill(0,a.size()){|idx| 
  if idx % 2 == 0
    a_rev[idx]
  else 
    a[idx]
  end
  }

def longest_num_start_idx(ary)
  start_idx = 0
  end_idx = 0
  seq_num = 0
  longest_start_idx = 0
  (ary.size()-1).times do |idx|
    if ary[idx] != ary[idx+1]
      end_idx = idx
      if seq_num < end_idx - start_idx + 1
        seq_num = end_idx - start_idx + 1
        longest_start_idx = start_idx
      end
    else
      start_idx = idx
    end
  end
  if seq_num == 0
    seq_num = ary.size()
  end
  return seq_num, longest_start_idx
end



seq_1 = longest_num_start_idx(b1)
seq_2 =  longest_num_start_idx(b2)

if(seq_1[0] == b1.size() && seq_2[0] == b2.size())
  p 0
else
  p a.count{|item| item != a[seq_1[1]] && item != a[seq_2[1]]}
end