CODEWITHSUNDEEP
arraysrubystring
Amrit Dhungana
Amrit Dhungana

Reputation: 4485

Ruby array conversion

I have a string of digits:

s = "12345678910"

As you can see it is the numbers 1 through 10 listed in increasing order. I want to convert it to an array of those numbers:

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

How can I do it?

Upvotes: 0

Views: 132

Answers (3)

Cary Swoveland
Cary Swoveland

Reputation: 110755

Assumptions

  • the first (natural) number extracted from the string is the first character of the string converted to an integer;
  • if the number n is extracted from the string, the next number extracted, m, satisfies n <= m (i.e., the sequence is monotonically non-decreasing);
  • if n is extracted from the string, the next number extracted will have as few digits as possible (i.e., at most one greater than the number of digits in n); and
  • there is no need to check the validity of the string (e.g., "54632" is invalid).

Code

def split_it(str)
  return [] if str.empty?
  a = [str[0]]
  offset = 1
  while offset < str.size
    sz = a.last.size
    sz +=1 if str[offset,sz] < a.last
    a << str[offset, sz]
    offset += sz
  end
  a.map(&:to_i)
end

Examples

split_it("12345678910")
  #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

split_it("12343636412252891407189118901")
  #=> [1, 2, 3, 4, 36, 36, 41, 225, 289, 1407, 1891, 18901]

Upvotes: 0

seph
seph

Reputation: 6076

Assuming a monotonic sequence, here's my run at it.

input = a.first.chars
output = []
previous_int = 0

until input.empty?
  temp = []
  temp << input.shift until temp.join.to_i > previous_int
  previous_int = temp.join.to_i
  output << previous_int
end

puts output.to_s

#=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Upvotes: 1

Datt
Datt

Reputation: 881

How about this:

a = ["123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899"]
b = a.first.each_char.map {|n| n.to_i }
if b.size > 8
  c = b[0..8]
  c += b[9..b.size].each_slice(2).map(&:join).map(&:to_i)
end

# It would yield as follows:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]

For later numbers beyond 99, modify existing predicate accordingly.

Upvotes: 5

Related Questions