Snow Leopards BG #47
Snow Leopards BG #47
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bukunmig
commented
Dec 9, 2022
bukunmig
commented
- Reviewed tests and can't see exceptions. Not sure why test is failing.
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Your submitted waves look good. Keep the algorithmic complexity in mind =, since just like python, many built-in JS array methods are hiding loops that we need to be aware of and may want to avoid when we're already within another loop. Please be sure to ping me when you've re-submitted with the revisions for the other waves so that I can review those as well.
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| it("returns a score of 0 if given an empty input", () => { | ||
| throw "Complete test"; | ||
| expectScores ({ |
| const correct = { word: "XXXX", score: scoreWord("XXXX") }; | ||
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| throw "Complete test by adding an assertion"; | ||
| expect(highestScoreFrom(words)).toEqual(correct); |
| let availableLetters = []; | ||
| let hand = []; | ||
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| for (const [key, value] of Object.entries(letterPool)) { |
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Consider using names more descriptive of what the key and value represent, such as letter and count.
src/adagrams.js
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| let letter = availableLetters[Math.floor(Math.random() * availableLetters.length)]; | ||
| hand.push(letter); | ||
| let index = availableLetters.indexOf(letter); |
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Alternatively, we could organize it this way and avoid needing to find the index
let index = Math.floor(Math.random() * availableLetters.length);
let letter = availableLetters[index];
hand.push(letter);| let letter = availableLetters[Math.floor(Math.random() * availableLetters.length)]; | ||
| hand.push(letter); | ||
| let index = availableLetters.indexOf(letter); | ||
| availableLetters.splice(index, 1); |
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Note that this has the same time costs as remove in python. That is, anything after the data that was removed needs to be shifted to fill the gap in the array, resulting in a O(n) time complexity with respect to the list length. Since we're working with fixed lenths (26 letters a certain number of times each, and a 10 letter hand), we dont need to worry too much about the absolute performance, but in general we want to find approaches that don't perform linear complexity operations within a loop.
src/adagrams.js
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| export const usesAvailableLetters = (input, lettersInHand) => { | ||
| // Implement this method for wave 2 | ||
| }; | ||
| // wave 2 |
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👀 Let me know when you've completed your rework on wave 2
src/adagrams.js
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| export const highestScoreFrom = (words) => { | ||
| // Implement this method for wave 4 | ||
| }; | ||
| // TODO: REDO Simple option |
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👀 Let me know when you've completed your re-work on wave 4
| // wave 2 | ||
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| export const scoreWord = (word) => { |
| // - Returns `true` if every letter in the `input` word is available (in the right quantities) in the `lettersInHand` | ||
| // - Returns `false` if not; if there is a letter in `input` that is not present in the `lettersInHand` or has too much of compared to the `lettersInHand` | ||
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| lettersInHand = [...lettersInHand]; |
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👍 Since we're splicing values out of the hand, we need this copy so that we don't destroy the passed in hand.
| if(lettersInHand.includes(letter)) { | ||
| let indexLettersInHand = lettersInHand.indexOf(letter); | ||
| lettersInHand.splice(indexLettersInHand, 1); |
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Notice that we can use the result from indexOf itself to tell us whether the value is in the array, since it returns -1 if it is not. So we could wrrite this as
const indexLettersInHand = lettersInHand.indexOf(letter);
if (indexLettersInHand !== -1) {
lettersInHand.splice(indexLettersInHand, 1);There was a problem hiding this comment.
This approcah (removing the values from from the hand copy as we use them) is going to be a quadratic O(n^2) approach (since the splice, like a remove, must shift everything else forward as things get removes). We could still think about building a frequency table of letters to determine whether the word can be made with the hand in linear time.
| winningWordAndScore ['score'] = value; | ||
| }; | ||
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| if (value === winningWordAndScore['score'] && winningWordAndScore['word'].length < 10) { |
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The tie breaker conditions can be an else if, since if the new score beat the old score outright, there's no need to check any of the tie breaking code.
| score: Object.values(wordsAndScores)[0] | ||
| }; | ||
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| for (const[key, value] of Object.entries(wordsAndScores)) { |
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Consider using more descriptive names for the key and value (maybe word and score) so we don't have to remember that the key is the word and the value is the score.
| // Implement this method for wave 3 | ||
| // wave 3 | ||
| let score = 0; | ||
| for (let letter of word) { |
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Prefer to use const for the loop control variable (letter).
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