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
v = gets.chomp.split.map(&:to_i)

def fetch_max(h, exclude_num: -1) 
  cnt = 0
  num = 0  # Sentinel
  h.each do |k, v|
    if k == exclude_num
      next
    end
    if v > cnt
      cnt = v
      num = k
    end
  end

  [cnt, num]
end

even = v.select.with_index { |_, i| i % 2 == 0 }
odd  = v.select.with_index { |_, i| i % 2 == 1 }

# Calculate max element of even
even_h = {}
even.each do |e|
  even_h[e] ||= 0
  even_h[e] += 1
end
max_even_cnt, max_even_num = fetch_max(even_h)

p "even num: #{max_even_num}, even cnt: #{max_even_cnt}"

# Calculate max element of odd
odd_h = {}
odd.each do |e|
  odd_h[e] ||= 0
  odd_h[e] += 1
end
max_odd_cnt, max_odd_num = fetch_max(odd_h)

p "odd num: #{max_odd_num}, odd cnt: #{max_odd_cnt}"

if max_even_num == max_odd_num
  if max_odd_cnt > max_even_cnt  # Use odd for this number
    max_even_cnt, max_even_num = fetch_max(even_h, exclude_num: max_odd_num)
  elsif max_odd_cnt < max_even_cnt
    max_odd_cnt, max_odd_num = fetch_max(odd_h, exclude_num: max_even_num)
  else # max_odd_cnt == max_even_cnt
    max_num = max_odd_num
    # Now, we have to use the largest next one
    even_cnt, even_num = fetch_max(even_h, exclude_num: max_num)
    odd_cnt,  odd_num  = fetch_max(odd_h, exclude_num: max_num)
    if even_cnt > odd_cnt  # even should be used as next
      max_even_cnt = even_cnt
      max_even_num = even_num
    else
      max_odd_cnt = odd_cnt
      max_odd_num = odd_num
    end
  end
end

p "even num: #{max_even_num}, even cnt: #{max_even_cnt}"
p "odd num: #{max_odd_num}, odd cnt: #{max_odd_cnt}"

p (n - max_odd_cnt - max_even_cnt)