Kay and Laura.rb#21
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…ictionary word, failing test that previously was passed in wave 4--testing merge previously passing code with current buggy code
AdagramsWhat We're Looking For
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CheezItMan
reviewed
Aug 23, 2018
| def draw_letters | ||
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| # all letters | ||
| letters = %W(a a a a a a a a a b b c c d d d d e e e e e e e e e e e e f f g g g h h i i i i i i i i i j k l l l l m m n n n n n n o o o o o o o o p p q r r r r r r s s s s t t t t t t u u u u v v w w x y y z ) |
| 10.times do | ||
| # check if index is in indices_used array , if true generate new random index | ||
| while indices_used.include?(index) | ||
| index = Random.rand(min..max) |
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Just a note, this loop could potentially go a long time. You would do better to remove an element once you've selected it.
| result << true | ||
| else | ||
| # user used a letter that does not exists | ||
| result << false |
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Maybe you should just return false.
| input.each_char do |letter| | ||
| # check if letter exists in letters_in_han | ||
| if letters_in_hand.include?(letter) | ||
| result << true |
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What if a letter appears twice in input, but not in letters_in_hand
| end | ||
| final_results << result.all? { |value| value == true} | ||
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| input.each_char do |letter| |
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This seems to be doing what the above did, but better. Avoid duplication of effort.
| group4 = %w(F H V W Y) | ||
| group5 = %w(J X) | ||
| group6 = %w(Q Z) | ||
| word.upcase.each_char do |letter| |
| score = data.select{ |k, v| k == winner}.values[0] | ||
| return {:word => winner, :score => score} | ||
| else # select shortest word and return it as a winner | ||
| winner = ties_words.sort_by {|x| x.length}.first |
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You have a loop and inside that you're sorting each iteration. This is a bit inefficient.
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Adagrams
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Comprehension Questions
Enumerablemixin? If so, where and why was it helpful?