Leaves - Julia K #27
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Leaves - Julia K #27
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added 10 commits
November 15, 2019 13:13
…idea how to pull letters from a JavaScript object -- hey, it's only the first week.
…o count letters drawn.
…t of a letter than is in letter pool. Added this logic to function. Passing wave 1 tests.
… containing 2 Zs. Now it is.
…hat tallies the count of each letter.
… return values in the right place.
| // code for random courtesy of https://stackoverflow.com/questions/4550505/getting-a-random-value-from-a-javascript-array | ||
| // pull a random letter from the letter pool array | ||
| let randomLetter = letterPoolArray[Math.floor(Math.random() * letterPoolArray.length)]; | ||
| // condition: if adding the letter to the player hand would exceed the count of letters in the pool, redraw |
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yes! nice work on this logic and clear comments describing what's going on.
| // Implement this method for wave 1 | ||
| const letterPoolHash = {'A':9, 'B':2, 'C':2, 'D':4, 'E':12, 'F':2, 'G':3, 'H':2, 'I':9, 'J':1, 'K':1, 'L':4, 'N':6, 'O':8, 'P':2, 'Q':1, 'R':6, 'S':4, 'T':6,'U':4, 'V':2, 'W':2, 'X':1, 'Y':2, 'Z':1 }; | ||
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| const letterPoolArray = ['A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'B', 'B', |
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This data structure works, but it's not very DRY and would be tricky to change. For example, suppose I said you had the wrong number of "I"s - you would have to do a lot of counting to solve the problem.
Instead you could use your hash of letter frequencies to create this array dynamically.
| } | ||
| }); | ||
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| // loop over handHash: if letter count in input is ever greater than letter count in hash, return false. Else return true. |
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Consider whether it would have been possible to do this check in the for loop above (had that not been a forEach loop)?
| let wordArray = word.toUpperCase().split('') | ||
| let score = 0 | ||
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| if (wordArray.length >= 7) { |
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you forgot && wordArray.length <= 10)
JS AdagramsWhat We're Looking For
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beccaelenzil
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Jun 2, 2022
…-5.2.3 Bump ws from 5.2.2 to 5.2.3
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