Tigers - Rose and Kindra #63
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| 'Z': 10 | ||
| } | ||
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| def create_letter_pool_list(LETTER_POOL): |
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Love a good helper function!! 👍 By creating a list of all letters by quantity, we get a more accurate hand of letter from the pool!
| count = hand_of_ten_letters.count(random_letter) | ||
| if count < LETTER_POOL[random_letter]: |
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Since the letter_pool_list has all the letters that can be pulled inside it, instead of finding the count of the letter every time, what if we just used remove and took out the letter from the list.
Now, the next time line 71 executes, it will have an accurate pool of letters that never need recounting. If it's in the list still, then it's fair game.
Either works! I don't think my suggestion is any more efficient than yours, however!
| if word.count(char) is not hand_of_ten_letters.count(char): | ||
| return False |
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Nice job! While the tests all passed, I think we can expand on the logic here to make it more robust. Let's say our word is "DOG" and our hand is [D,G,G,I,O,D,L,A,S,T]
Would your function return this as true or false?
| for char in word: | ||
| if word.count(char) is not hand_of_ten_letters.count(char): | ||
| return False | ||
| return True | ||
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| def score_word(word): |
| return(highest_score_tuple) |
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We could combine these together in one line:
return highest_scoring_word,highest_score
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