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Discussion on: Project Euler #3 - Largest Prime Factor

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hanachin profile image
Seiei Miyagi

Ruby✨💎✨

require "prime"

puts Enumerator.new { |y|
  n = 600851475143

  Prime.each do |prime|
    break if n == 1

    q, r = n.divmod(prime)

    next if r.nonzero?

    y << prime
    n = q
    redo
  end
}.to_a.last
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ezeilosu profile image
Sunday Ezeilo

Nice one. But, I think y << prime, to_a.last pose addition execution time. Michael Kohl's approach is more optimized.