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re: Challenge: find 'Kaprekar numbers' VIEW POST

FULL DISCUSSION
 

Here's my Ruby novice solution -- Works for the known oddballs (4879, 5292, 38962). Prepending '0' to odd-sized numbers inspired by others here.

def kaprekar?(k)
  s = k.to_s.size    # s = nr. of digits in k
  k_sq = (k**2).to_s    # k_sq = square of k (String)
  k_sq.prepend("0") if (k_sq.size.even? == false)    # prepend '0' to odd-sized numbers
  if k_sq[s-1] == "0"    # if digit 's-1' is 0, loop different combinations
    (0..s).each do |x|
      part1 = k_sq[0..x-1].to_i
      part2 = k_sq[x..-1].to_i
      return true if (k == part1 + part2)
    end
  else    # if not, split before digit 's' (Integers)
    part1 = k_sq[0..s-1].to_i
    part2 = k_sq[s..-1].to_i
    return true if k == (part1 + part2)
  end
  false    # else returns false
end
 

I don't like the exceptions (4879, 5292, 38962...) -- it feels like they don't belong in the sequence. And look how pretty the code is without them!

  def kaprekar?(k)
    s = k.to_s.size
    k_sq = (k**2).to_s
    k_sq.prepend("0") if (k_sq.size.even? == false)
    part1 = k_sq[0..s-1].to_i
    part2 = k_sq[s..-1].to_i
    return true ? k == part1 + part2 : false
  end
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