Regular expression: if three letters are included

I've got a number of word: hello, poison, world, search, echo ... And I've got some letters e, h, o Now I need to find all words that includes this letters. Like search, echo for e, h, o

I can search this way:

words = %w[hello poison world search echo]
matched = words.select do |w|
  %w[e,h,o].all?{ |l| w =~ /#{l}/ }
end

The problem is that if letters are o, o, o, or l, b, l this search will return true for words like open or boil, but 开发者_运维百科I need to search words that includes three of o or two of l and one b

upd:

leters = "abc"
words.select{ |w| w.count(letters) >= 3 }

upd 2

Bad solution, example:

"lllllll".count("lua") #=> 5

Are you sure you want a regexp? Strings support counting values in them, you could use that feature. Something like this:

words = ["pool", "tool", "troll", "lot"]
letters = "olo"

#find how many of each letter we need
counts = {}
letters.each { |v| counts[v] = letters.count(v) }

#See if a given work matches all the counts
# accumulated above
res = words.select do |w|
  counts.keys.inject(true) do |match, letter| 
      match && (w.count(letter) == counts[letter])
  end
end

It's probably best not to use regular expressions for this, but it can be done:

All three letters different:

/^(?=.*a)(?=.*b).*c/

Two the same and one different:

/^(?=.*a.*a).*b/

All three the same:

/^.*a.*a.*a/

Consider modifying the word (so that it becomes smaller with each check).

words = %w(fooo for find o ooo)
matched = words.select do |orig|
  # note: str.gsub! returns nil if nothing was replaced
  w = orig.clone
  %w(o o o).all?{ |l| w.gsub!(/^(.*)(#{l})(.*)$/, '\1\3') }
end

Looks crazy, but it works:

leters = "abc"
words.select{ |w| w.count(letters) >= 3 }

But it isn't work with cyrillic letters :(