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.to_i
nums = gets.split.map(&:to_i)
 
evens = []
odds = []
 
count = 0
nums.each do |num|
  if count.even?
    evens.push(num)
  else
    odds.push(num)
  end
  count += 1
end
 
evenFreq = evens.each_with_object(Hash.new(0)){ |m,h| h[m] += 1 }.sort_by{ |k,v| v }.reverse
oddFreq =  odds.each_with_object(Hash.new(0)){ |m,h| h[m] += 1 }.sort_by{ |k,v| v }.reverse
 
#p evenFreq
#p oddFreq

if evenFreq == oddFreq
  res = 0
  evenFreq.each do |num|
    res += num[1]
  end
  puts res
  exit 0
end
 
if evenFreq.size == 1 && oddFreq.size == 1
  puts 0
  exit 0
end
 
@evenMost = evenFreq[0][0]
@oddMost = oddFreq[0][0]
 
if @evenMost == @oddMost
  if evenFreq[0][1] > oddFreq[0][1]
    @oddMost = oddFreq[1][0]
  else
    @evenMost = evenFreq[1][0]
  end
end
 
count = 0
evenFreq.each do
  if evenFreq[count][0] == @evenMost
    evenFreq.delete_at(count)
    break
  end
  count += 1
end
 
count = 0
oddFreq.each do
  if oddFreq[count][0] == @oddMost
    oddFreq.delete_at(count)
    break
  end
  count += 1
end
 

#p evenFreq
#p oddFreq

res = 0
evenFreq.each do |num|
  res += num[1]
end
oddFreq.each do |num|
  res += num[1]
end
 
puts res