adagrams #4
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CheezItMan
reviewed
Sep 23, 2020
| @@ -0,0 +1,139 @@ | |||
| # frozen_string_literal: true | |||
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| a = Array.new(9) { 'A' } | |||
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This method works, but it's pretty long-hand.
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| ALPHABET = [a, b, c, d, e, f, g, h, i, j, k, l, m, n, | ||
| o, p, q, r, s, t, u, v, w, x, y, z].flatten | ||
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| SCORE_CHART = [ | ||
| { 1 => %w[A E I O U L N R S T] }, | ||
| { 2 => %w[D G] }, | ||
| { 3 => %w[B C M P] }, | ||
| { 4 => %w[F H V W Y] }, | ||
| { 5 => %w[K] }, | ||
| { 8 => %w[J X] }, | ||
| { 10 => %w[Q Z] } | ||
| ].freeze |
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I like how you included these constants at the top for reference.
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| # wave 1 | ||
| def draw_letters | ||
| ten_letters = ALPHABET.sample(10) | ||
| return ten_letters | ||
| end |
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| count = letter_frequency[letter] | ||
| count += 1 | ||
| letter_frequency[letter] = count |
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A little simplification.
Suggested change
| count = letter_frequency[letter] | |
| count += 1 | |
| letter_frequency[letter] = count | |
| letter_frequency[letter] += 1 |
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| count = letter_count[letter] | ||
| if count == 0 | ||
| false | ||
| else | ||
| count -= 1 | ||
| letter_count[letter] = count | ||
| true | ||
| end |
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This works, but you can simplify it a bit.
Suggested change
| count = letter_count[letter] | |
| if count == 0 | |
| false | |
| else | |
| count -= 1 | |
| letter_count[letter] = count | |
| true | |
| end | |
| letter_count[letter] -= 1 | |
| letter_count[letter] < 0 |
| end | ||
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| # wave 3 | ||
| def score_word(word) |
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👍
Really clever use of select, and include?.
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| # wave 4 | ||
| def highest_score_from(words) |
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Assignment Submission: Adagrams
Congratulations! You're submitting your assignment. Please reflect on the assignment with these questions.
Reflection
Enumerablemixin? If so, where and why was it helpful?