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: Time Limit Exceeded)

items_count = gets
items = gets.split(" ")
items_odd = items.select.with_index {|item,index| index.odd?}
items_even = items.select.with_index {|item,index| index.even?}

count_odd = 0
count_even = 0

length = items.length 
old_count = 0

pick_up_odd_int = 0
pick_up_even_int = 0

length.times do |l|
    int = items_odd[l].to_i
    old_count = items_odd.count(items_odd[l]).to_i
    if old_count > count_odd 
        pick_up_odd_int = int
        count_odd = old_count 
    end 
end 

length.times do |l|
    int = items_even[l].to_i
    old_count = items_even.count(items_even[l]).to_i
    if old_count > count_even 
        pick_up_even_int = int
        count_even = old_count 
    end 
end 

result = 0

unless pick_up_even_int == pick_up_odd_int
    result = (length / 2 - count_odd) + (length / 2 - count_even)

else 
    count_even = length / 2 - count_even 
    result = (length / 2 - count_odd) + (length / 2 - count_even)

end 

p result.abs