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)

def frequent_elements(vs)
  tbl = {}
  vs.each do |v|
    tbl[v] ||= 0
    tbl[v] += 1
  end
  tbl.to_a.sort_by(&:last).map(&:first).reverse
end

n = gets.to_i
vs = gets.split.map(&:to_i)

odds, evens = vs.each_slice(2).to_a.transpose

oe = frequent_elements(odds)
ee = frequent_elements(evens)

if oe[0] == ee[0]
  if oe.size == 1 && ee.size == 1
    puts n/2
  elsif oe.size == 1
    puts n/2 - evens.count(ee[1])
  elsif ee.size == 1
    puts n/2 - odds.count(oe[1])
  end
else
  puts n - evens.count(ee[0]) - odds.count(oe[0])
end