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Leaves - Raisah #46

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Leaves - Raisah #46

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@raisahv raisahv commented Nov 18, 2019

JS Adagrams

Congratulations! You're submitting your assignment!

Comprehension Questions

Question Answer
What patterns were you able to use from your Ruby knowledge to apply to JavaScript? I was able to apply the same debugging strategies that I used from Ruby to JavaScript, though with JavaScript I had to focus more on syntax and reviewing my code before expecting it to work.
Did you need to use different strategies to find information online about JavaScript than you would for Ruby? No, I generally used the same strategies to find information online about JavaScript.
What was a challenge you were able to overcome on this assignment? I was excited to convert the Adagrams to a class and reformat the tests to use the JS version of a before loop to DRY up the code.
What has been interesting and positive about learning a new programming language? I felt more confident learning JavaScript since it went so much faster than learning Ruby.
What is something to focus on your learnings in JavaScript in the next week? Learning more about how to use classes.

@dHelmgren
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JS Adagrams

What We're Looking For

Feature Feedback
General
Answered comprehension questions yes
Small commits with meaningful commit messages yes
Code Requirements
drawLetters method good
Uses appropriate data structure to store the letter distribution yes
All tests for drawLetters pass yes
usesAvailableLetters method good
All tests for usesAvailableLetters pass yes
scoreWord method good
Uses appropriate data structure to store the letter scores yes
All tests for scoreWord pass yes
Optional
highestScoreFrom method good
Appropriately handles edge cases for tie-breaking logic yes
All tests for highestScoreFrom pass yes
Overall Great job overall! This code is impeccably organized and well thought out! Keep up the hard work!

},
};
const letterPoolObject = {
A: 9,

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Nice work using a JS Object here!

const handIndex = lettersInHand.indexOf(input[charIndex]);
lettersInHand.splice(handIndex, 1);
} else {
allValid = false;

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I believe you could short circuit out of this for by returning here, not that it will save you much time in this loop.

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2 participants