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)

a_nums = []
b_nums = []

index = 0
nums.each do |num|
  if index.even?
    a_nums.push(num)
  else
    b_nums.push(num)
  end
  index += 1
end

def getFreqList(nums)
  freqList = Hash.new
  nums.each do |num|
    if freqList[num].nil?
      freqList[num] = 1
    else
      freqList[num] += 1
    end
  end
  newList = []
  freqList.sort_by{|k, v| v}.reverse.each do |item|
    item[1].times do
      newList.push(item[0])
    end
  end
  newList
end

# p getFreqList(nums)

a_freqList = getFreqList(a_nums)#a_nums.sort_by!{|num| a_nums.count num}.reverse! #100000000
b_freqList = getFreqList(b_nums)#b_nums.sort_by!{|num| b_nums.count num}.reverse!

a_count = 0
b_count = 0

while a_freqList[a_count] == b_freqList[b_count]
  if b_count == a_count
    a_count += 1
  else
    b_count += 1
  end
end

res = 0

res += a_nums.select {|num| num != a_freqList[a_count]}.count + b_nums.select {|num| num != b_freqList[b_count]}.count #20000
p res