Lions - Suzanne S18 #42
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| @@ -145,7 +147,7 @@ describe("Adagrams", () => { | |||
| const words = ["XXX", "XXXX", "X", "XX"]; | |||
| const correct = { word: "XXXX", score: scoreWord("XXXX") }; | |||
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| throw "Complete test by adding an assertion"; | |||
| expect(highestScoreFrom(words)).toEqual(correct); | |||
| expectScores({ | ||
| "": 0, | ||
| }); |
| @@ -1,15 +1,126 @@ | |||
| export const drawLetters = () => { | |||
| // Implement this method for wave 1 | |||
| const letterBank = { | |||
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It would be good to have this declared outside of the function so it doesn't have to be recreated each time the function is called.
| Y: 2, | ||
| Z: 1, | ||
| }; | ||
| function getRandomLetter(letterBank) { |
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It would be good to have this declared outside of the function so it doesn't have to be recreated each time the function is called.
| const hand = []; | ||
| while (hand.length < 10) { | ||
| const randomLetter = getRandomLetter(letterBank); | ||
| const occurance = hand.filter((x) => x === randomLetter).length; | ||
| if (occurance < letterBank[randomLetter]) { | ||
| hand.push(randomLetter); | ||
| } | ||
| } | ||
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| return hand; |
| for (let inputLetter of input) { | ||
| const index = drawn.indexOf(inputLetter); | ||
| if (index > -1) { | ||
| drawn.splice(index, 1); |
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We should not be removing letters from the drawn array. All this function should do is check to see if all of the letters in input are in drawn. If letters are being removed every time we're just checking to see if the input word is an anagram, we'll run out of letters pretty quickly.
| }; | ||
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| export const scoreWord = (word) => { | ||
| // Implement this method for wave 3 | ||
| const scoreChart = { |
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It would be good to declare this outside of the function.
| for (let letter of word.toUpperCase()) { | ||
| let key = Object.keys(scoreChart).filter(function (key) { | ||
| return key.includes(letter); | ||
| }); | ||
| const score = scoreChart[key]; | ||
| total += score; | ||
| } |
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It would be more efficient to store the scores with one letter to one number value. Looking up a single key in a JS object is O(1) time complexity. Here, you need to look through every letter to find the one that matches.
| { | ||
| break; | ||
| } |
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Since we know this is definitely the best word, we can just do return wordObj from here and avoid the logic at the bottom looking for the object in the array again.
| break; | ||
| } | ||
| } else if (wordObj.word.length < minLength) { | ||
| bestWord = wordObj.word; |
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Instead of keeping track of just the best word, we can keep track of the whole wordObj that's the current best. This will allow us to avoid the logic later of trying to find the object again.
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