CodeWars (5 kyu) Pete, the baker - Javascript

Description:

Pete likes to bake some cakes. He has some recipes and ingredients. Unfortunately he is not good in maths. Can you help him to find out, how many cakes he could bake considering his recipes?

Write a function cakes(), which takes the recipe (object) and the available ingredients (also an object) and returns the maximum number of cakes Pete can bake (integer). For simplicity there are no units for the amounts (e.g. 1 lb of flour or 200 g of sugar are simply 1 or 200). Ingredients that are not present in the objects, can be considered as 0.

Examples:

// must return 2
cakes({flour: 500, sugar: 200, eggs: 1}, {flour: 1200, sugar: 1200, eggs: 5, milk: 200}); 
// must return 0
cakes({apples: 3, flour: 300, sugar: 150, milk: 100, oil: 100}, {sugar: 500, flour: 2000, milk: 2000});

Sample Tests

const {assert} = require('chai');

describe('description example', function() {
  it('pass example tests', function() {
    let recipe = {flour: 500, sugar: 200, eggs: 1};
    let available = {flour: 1200, sugar: 1200, eggs: 5, milk: 200};
    assert.equal(cakes(recipe, available), 2);

    recipe = {apples: 3, flour: 300, sugar: 150, milk: 100, oil: 100};
    available = {sugar: 500, flour: 2000, milk: 2000};
    assert.equal(cakes(recipe, available), 0);
  });
});

Solution

To solve this kata, we'll be taking in recipe, which contains the required ingredients and their respective amounts stored within an object. Additionally, we have another object available that tells us what ingredients we have on hand.

We'll need to see what ingredients the recipe calls for and figure out how many cakes we can make with the ingredients we have in available. If recipe calls for an ingredient that does not exist in available, we'll need to return 0.

First, let's work through the pseudocode:

function cakes(recipe, available) {
    // Iterate over recipe object to get each ingredient
    // Confirm we have the ingredient available
    // Calculate the maximum amount we can make per each ingredient

    // Return the smallest amount we can make
}

Now that we have an idea of how to proceed, let's start working through the coding solution.

function cakes(recipe, available) {
    // Need a variable to keep track of how many possible cakes can be made per ingredient
    let counterArr = []

    // Iterate over `recipe` object to get each ingredient
    for (let ingredient in recipe){
    // Confirm we have the ingredient available
    // Calculate the maximum amount we can make per each ingredient
            available[ingredient] ?     
counterArr.push(Math.floor(available[ingredient] / recipe[ingredient])) : counterArr.push(0)
    }

    // Return the smallest amount we can make
    return Math.min(...counterArr)
}

So there are a few pieces in the code we should make note of:

1. We'll need one additional variable counterArr to keep track of how many possible cakes we can make per ingredient

2. We'll be iterating over the recipe object by using for...in

3. We check if we have the recipe's required ingredients available

   1. If available, we push the maximum, integer amount of cakes we can make per ingredient

   2. If not available, then we can't make any cakes and push 0

4. We return the smallest number in our counterArr by combining Math.min() with the spread operator ...