二维数组第二元素上的Ruby插值搜索

Question

What I have below is a simple interpolation search on a single dimensional array $employee_list . The list is an ordered list of employees' id that might have gaps due to retirements.

def exist?(id)
  lower = 0
  upper = $employee_list.length - 1
  while $employee_list[upper] != $employee_list[lower] && id >= $employee_list[lower] && id <= $employee_list[upper]
      middle = lower + ((id - $employee_list[lower]) * (upper - lower) / ($employee_list[upper] - $employee_list[lower]))
      if id > $employee_list[middle]
        lower = middle + 1
      elsif id < $employee_list[middle]
        upper = middle - 1
      else
        return true
      end
    end
  return false
end

Now I want to add a new element to the list and the 2nd element of the array will contain the employee birth year (ie $employee_list[id][birthyear] ). I am able to sort the array based on the birthyear and I would like to perform a interpolation search based on the birthyear and return the list of employee ids that have that particular birthyear.

Answer 1

Given the following employee_list :

employee_list = [[19, 1992], [41, 1985], [12, 1958], [63, 1985]]

If you want to get the ids of the employees born in 1985 you just need to select those arrays whose last element is 1985, and get the first element of the filtered arrays:

employee_list.select{|employee| employee.last == 1985}.map(&:first)
# => [41, 63]

Answer 2

Here's a modified version. It returns one id if an employee with given year has been found, and nil if no employee has been found.

You'll need a modified version of interpolation search if you need to return multiple ids for a year. Finally, interpolation search works best with uniformly distributed values. I guess it's really not the case for birthyear of employees.

def find_id_for_year(array, year)
  lower = 0
  upper = array.length - 1
  while array[upper][1] != array[lower][1] && year >= array[lower][1]  && year <= array[upper][1]
    middle = lower + ((year - array[lower][1]) * (upper - lower) / (array[upper][1] - array[lower][1]))
    if year > array[middle][1]
      lower = middle + 1
    elsif year < array[middle][1]
      upper = middle - 1
    else
      return array[middle][0]
    end
  end
end

employee_list = Array.new(10) {|i| [i, rand(1990..2000)] }.sort_by(&:last)

p employee_list
#=> [[5, 1990], [8, 1990], [2, 1991], [9, 1991], [7, 1992], [0, 1995], [6, 1996], [4, 1998], [1, 1999], [3, 1999]]
#=> [[8, 1990], [4, 1991], [5, 1991], [6, 1991], [2, 1996], [0, 1998], [1, 1998], [3, 1998], [7, 1999], [9, 2000]]

p find_id_for_year(employee_list, 1992)
#=> 7
#=> nil