Tigers - Mica C. and Lain #64
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| Original file line number | Diff line number | Diff line change |
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| @@ -1,11 +1,89 @@ | ||
| LETTER_POOL = { | ||
| "A": 9, | ||
| "N": 6, | ||
| "B": 2, | ||
| "O": 8, | ||
| "C": 2, | ||
| "P": 2, | ||
| "D": 4, | ||
| "Q": 1, | ||
| "E": 12, | ||
| "R": 6, | ||
| "F": 2, | ||
| "S": 4, | ||
| "G": 3, | ||
| "T": 6, | ||
| "H": 2, | ||
| "U": 4, | ||
| "I": 9, | ||
| "V": 2, | ||
| "J": 1, | ||
| "W": 2, | ||
| "K": 1, | ||
| "X": 1, | ||
| "L": 4, | ||
| "Y": 2, | ||
| "M": 2, | ||
| "Z": 1, | ||
| } | ||
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| LETTER_POINTS = { | ||
| "A": 1, | ||
| "N": 1, | ||
| "B": 3, | ||
| "O": 1, | ||
| "C": 3, | ||
| "P": 3, | ||
| "D": 2, | ||
| "Q": 10, | ||
| "E": 1, | ||
| "R": 1, | ||
| "F": 4, | ||
| "S": 1, | ||
| "G": 2, | ||
| "T": 1, | ||
| "H": 4, | ||
| "U": 1, | ||
| "I": 1, | ||
| "V": 4, | ||
| "J": 8, | ||
| "W": 4, | ||
| "K": 5, | ||
| "X": 8, | ||
| "L": 1, | ||
| "Y": 4, | ||
| "M": 3, | ||
| "Z": 10, | ||
| } | ||
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| def draw_letters(): | ||
| from random import sample | ||
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| pool = "".join(letter * count for letter, count in LETTER_POOL.items()) | ||
| return sample(pool, k=10) | ||
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| def uses_available_letters(word, letter_bank): | ||
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There was a problem hiding this comment. This is a fine implementation very succinct and pythonic, though there's a slight time performance hit from using |
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| upper_word = word.upper() | ||
| return all( | ||
| upper_word.count(letter) <= letter_bank.count(letter) | ||
| for letter in set(upper_word) | ||
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There was a problem hiding this comment. From my investigations, A common way to implement this function is to create frequency dictionaries of either or both However, by creating the dictionaries, we are also using O(n) space, while the big benefit of generators is that they don't need to allocate space, making y'all's implementation O(1). Which could very well be worth the worse time complexity! Trade-offs abound. |
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| ) | ||
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| def score_word(word): | ||
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There was a problem hiding this comment. This is also a very fine succinct and pythonic implementation, but I think then we get a space performance issue this time... |
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| total_score = sum([LETTER_POINTS.get(letter, 0) for letter in word.upper()]) | ||
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There was a problem hiding this comment. In this case rather than a generator, y'all used a comprehension (due to the square brackets), which does end up allocating O(n) memory (apparently even if you never assign it, sadly). However, I just tested and y'all could just switch to a generator here and get O(1) space complexity! Great since this is so succinct and clean and I do really value that! |
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| total_score += len(word) > 6 and 8 | ||
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There was a problem hiding this comment. Good use of the short-circuiting behavior of |
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| return total_score | ||
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| def break_tie(words): | ||
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| return min(words, key=lambda w: 0 if len(w) == 10 else len(w)) | ||
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There was a problem hiding this comment. And this is so good and succinct to account for all tiebreakers and verifying that Nice work! |
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| def get_highest_word_score(word_list): | ||
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There was a problem hiding this comment. Nice and succinct as all the rest of the code has been while still being correct. No comments. |
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| word_scores = {word: score_word(word) for word in word_list} | ||
| highest_score = max(word_scores.values()) | ||
| tied_words = [word for word, score in word_scores.items() if score == highest_score] | ||
| return break_tie(tied_words), highest_score | ||
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This looks fine, good use of library functions and built-in operators! No comments.