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: Accepted)

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

odds = v.each_slice(2).map(&:first)
evens = v.each_slice(2).map(&:last)

h_odd = Hash.new(0)
h_even = Hash.new(0)

odds.each do |val|
  h_odd[val] += 1
end

evens.each do |val|
  h_even[val] += 1
end

h_odd = h_odd.sort_by{ |k, v| -v}
h_even = h_even.sort_by{ |k, v| -v}

# oddとevenの数字が同じなってしまった場合はあとで変更できる？
# それぞれの出現回数をとっておく
# それぞれ一番多いやつに合わせるのが正解
# もし同じときはどっちかが二番目 両方みてminにすれば良さそう

if h_odd[0][0] != h_even[0][0]
  puts (n/2 - h_odd[0][1]) + (n/2 - h_even[0][1])
elsif h_odd.size == 1 && h_even.size == 1
  puts n/2
elsif h_odd.size == 1
  puts n/2 - h_even[0][1]
elsif h_even.size == 1
  puts n/2 - h_odd[0][1]
else
  a = (n/2 - h_odd[0][1]) + (n/2 - h_even[1][1])
  b = (n/2 - h_odd[1][1]) + (n/2 - h_even[0][1])

  puts [a, b].min
end